Cambridge Additional Math : Coordinate Geometry of Circles

IGCSE O Level Mathematics

Coordinate Geometry of the Circle

Exercise 7.1

1. Find the centre and the radius of each of the following circles.
   (a) $ x^2+y^2=25$

   (b) $x^2+y^2=81$

   (c) $x^2+y^2=48$

   (d) $2 x^2+2 y^2=64$

   (e) $ x^2+(y-3)^2=1$

   (f) $(x-5)^2+y^2=9$

   (g) $(x+7)^2+y^2=18$

   (h) $(x+1)^2+(y-2)^2=100$

   (i) $(x-2)^2+(y+4)^2=9$

   (j) $(x+3)^2+(y+4)^2=27$
   $\begin{aligned} &\textbf{(a)} \quad x^2 + y^2 = 25 \quad \implies \text{centre: } (0, 0),~ \text{radius: } 5 \\ &\textbf{(b)} \quad x^2 + y^2 = 81 \quad \implies \text{centre: } (0, 0),~ \text{radius: } 9 \\ &\textbf{(c)} \quad x^2 + y^2 = 48 \quad \implies \text{centre: } (0, 0),~ \text{radius: } 4\sqrt{3} \\ &\textbf{(d)} \quad 2x^2 + 2y^2 = 64 \implies x^2 + y^2 = 32 \implies \text{centre: } (0, 0),~ \text{radius: } 4\sqrt{2} \\ &\textbf{(e)} \quad x^2 + (y - 3)^2 = 1 \quad \implies \text{centre: } (0, 3),~ \text{radius: } 1 \\ &\textbf{(f)} \quad (x - 5)^2 + y^2 = 9 \quad \implies \text{centre: } (5, 0),~ \text{radius: } 3 \\ &\textbf{(g)} \quad (x + 7)^2 + y^2 = 18 \quad \implies \text{centre: } (-7, 0),~ \text{radius: } 3\sqrt{2} \\ &\textbf{(h)} \quad (x + 1)^2 + (y - 2)^2 = 100 \quad \implies \text{centre: } (-1, 2),~ \text{radius: } 10 \\ &\textbf{(i)} \quad (x - 2)^2 + (y + 4)^2 = 9 \quad \implies \text{centre: } (2, -4),~ \text{radius: } 3 \\ &\textbf{(j)} \quad (x + 3)^2 + (y + 4)^2 = 27 \quad \implies \text{centre: } (-3, -4),~ \text{radius: } 3\sqrt{3} \end{aligned}$

2. Find the equation of each of the following circles.
   (a) centre $(0,0)$, radius 4

   (b) centre $(3,-1)$, radius 6

   (c) centre $(2,-5)$, radius $2 \sqrt{5}$

   (d) centre $\left(-\frac{1}{2},-\frac{3}{2}\right)$, radius 4
   $\begin{aligned} &\textbf{(a)} \quad \text{centre: } (0, 0),~ \text{radius: } 4 \implies x^2 + y^2 = 16 \\ &\textbf{(b)} \quad \text{centre: } (3, -1),~ \text{radius: } 6 \implies (x - 3)^2 + (y + 1)^2 = 36 \\ &\textbf{(c)} \quad \text{centre: } (2, -5),~ \text{radius: } 2\sqrt{5} \implies (x - 2)^2 + (y + 5)^2 = 20 \\ &\textbf{(d)} \quad \text{centre: } \left(-\frac{1}{2}, -\frac{3}{2}\right),~ \text{radius: } 4 \implies \left(x + \frac{1}{2}\right)^2 + \left(y + \frac{3}{2}\right)^2 = 16 \end{aligned}$

3. Find the centre and the radius of each of the following circles.
   (a) $x^2+y^2-4 x+10 y+25=0$

   (b) $x^2+y^2+14 x-8 y+40=0$

   (c) $x^2+y^2+4 x+12 y+4=0$

   (d) $x^2+y^2-2 x-16 y+33=0$

   (e) $ 4 x^2+4 y^2-4 x-16 y+1=0$

   (f) $4 x^2+4 y^2+24 x-12 y+13=0$
   $\begin{aligned} &\textbf{(a)} \quad x^2 + y^2 - 4x + 10y + 25 = 0 \\ & x^2 - 4x + y^2 + 10y = -25 \\ & x^2 - 4x + 2^2 + y^2 + 10y + 5^2 = -4 \\ & (x - 2)^2 + (y + 5)^2 = 4 \\ & \text{centre: } (2, -5), \text{ radius: } 2 \\[1em] &\textbf{(b)} \quad x^2 + y^2 + 14x - 8y + 40 = 0 \\ & x^2 + 14x + y^2 - 8y = -40 \\ & x^2 + 14x + 7^2 + y^2 - 8y + (-4)^2 = 25 \\ & (x + 7)^2 + (y - 4)^2 = 25 \\ & \text{centre: } (-7, 4), \text{ radius: } 5 \\[1em] &\textbf{(c)} \quad x^2 + y^2 + 4x + 12y + 4 = 0 \\ & x^2 + 4x + y^2 + 12y = -4 \\ & x^2 + 4x + 2^2 + y^2 + 12y + 6^2 = 36 \\ & (x + 2)^2 + (y + 6)^2 = 36 \\ & \text{centre: } (-2, -6), \text{ radius: } 6 \\[1em] &\textbf{(d)} \quad x^2 + y^2 - 2x - 16y + 33 = 0 \\ & x^2 - 2x + y^2 - 16y = -33 \\ & x^2 - 2x + 1^2 + y^2 - 16y + (-8)^2 = -33 + 1 + 64 \\ & (x - 1)^2 + (y - 8)^2 = 32 \\ & \text{centre: } (1, 8), \text{ radius: } 4\sqrt{2} \end{aligned}$ $\begin{aligned} &\textbf{(e)} \quad 4x^2 + 4y^2 - 4x - 16y + 1 = 0 \\ & x^2 + y^2 - x - 4y = -\frac{1}{4} \\ & x^2 - x + \left( \frac{1}{2} \right)^2 + y^2 - 4y + 4 = -\frac{1}{4} + \frac{1}{4} + 4 \\ & (x - \frac{1}{2})^2 + (y - 2)^2 = 4 \\ & \text{centre: } \left( \frac{1}{2}, 2 \right), \text{ radius: } 2 \\[1em] &\textbf{(f)} \quad 4x^2 + 4y^2 + 24x - 12y + 13 = 0 \\ & x^2 + y^2 + 6x - 3y = -\frac{13}{4} \\ & x^2 + 6x + 3^2 + y^2 - 3y + \left( \frac{3}{2} \right)^2 = -\frac{13}{4} + 9 + \frac{9}{4} \\ & (x + 3)^2 + \left( y - \frac{3}{2} \right)^2 = 8 \\ & \text{centre: } \left( -3, \frac{3}{2} \right), \text{ radius: } 2\sqrt{2} \end{aligned}$

4. Find the equation of the circle with centre $(3,2)$ passing through the point $(7,5)$.
   $\begin{aligned} &\text{centre: } (3, 2),\quad \text{point on circle: } (7, 5) \\ &\text{radius: } \sqrt{(7 - 3)^2 + (5 - 2)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \\ &\text{equation: } (x - 3)^2 + (y - 2)^2 = 25 \end{aligned}$

5. A diameter of a circle has its end points at $A(5,-3)$ and $B(15,5)$. Find the equation of the circle.
   $\begin{aligned} &\text{centre: } \left( \frac{5 + 15}{2}, \frac{-3 + 5}{2} \right) = (10, 1) \\ &\text{point on circle: } (5, -3) \\ &\text{radius: } \sqrt{(5 - 10)^2 + (-3 - 1)^2} = \sqrt{25 + 16} = \sqrt{41} \\ &\text{equation: } (x - 10)^2 + (y - 1)^2 = 41 \end{aligned}$

6. Sketch the circle $(x-4)^2+(y-5)^2=16$.

7. Find the equation of the circle that touches the $y$-axis and whose centre is $(-3,2)$.
   $\begin{aligned} &\text{centre: } (-3, 2) \\ &\text{point on circle: } (0, 2) \quad \text{(since the circle touches the } y\text{-axis)} \\ &\text{radius: } \sqrt{(0 - (-3))^2 + (2 - 2)^2} = \sqrt{9 + 0} = 3 \\ &\text{equation: } (x + 3)^2 + (y - 2)^2 = 9 \end{aligned}$

8. Show that $x^2+y^2+8 x-14 y=10$ can be written in the form $(x-a)^2+(y-b)^2=r^2$ where $a, b$ and $r$ are constants to be found. Hence, write down the coordinates of the centre of the circle and also the radius of the circle.
   $\begin{aligned} &x^2 + 8x + y^2 - 14y = 10 \\ &x^2 + 8x + 16 + y^2 - 14y + 49 = 10 + 16 + 49 \\ &(x + 4)^2 + (y - 7)^2 = 75 \\ &\text{centre: } (-4, 7) \\ &\text{radius: } \sqrt{75} = 5\sqrt{3} \end{aligned}$

9. The points $A(-9,-1)$ and $B(3,-5)$ lie on the circumference of a circle. Show that the centre of the circle lies on the line $y=3 x+6$.
   $\begin{aligned} &\text{Midpoint of } AB=\left(\frac{-9+3}{2}, \frac{-1-5}{2}\right)=(-3,-3) \\[4pt] &\text{Let $M$ be the midpoint of }AB.\\[4pt] &\text{Slope of } AB = \frac{-5 - (-1)}{\,3 - (-9)} = \frac{-4}{12} = -\frac13 \\[4pt] &\text{Slope of perpendicular bisector } = 3 \\[4pt] &\text{Equation through } M(-3,-3):\; y + 3 = 3(x + 3)\implies y = 3x + 6 \\[4pt] &\therefore\; \text{centre lies on } y = 3x + 6 \end{aligned}$ Alternative Solution Let $(a, b)$ be the centre of the circle. Hence the equation of the circle is $(x-a)^2+(y-b)^2=r^2$ $A(-9,-1)$ and $B(3,-5)$ lie on the circumference of the circle $\begin{aligned} \therefore\quad &(-9-a)^2+(-1-b)^2=(3-a)^2+(-5-b)^2 \\ & 81+18 a+a^2+1+2 b+b^2=9-6 a+a^2+25+10 b+b^2 \\ & 82+18 a+6 a-34=8 b \\ & 8 b=24 a+48 \\ & b=3 a+6 \end{aligned}$ $\therefore\ $ The centre of the circle lie on the line $y=3 x+6$.

10. Find the equation of the circle that passes through the point $(5,-2)$ and has the same centre as the circle $x^2+y^2-4 x+8 y-40=0$.
    $\begin{aligned} &x^2+y^2-4 x+8 y-40=0\\ & x^2-4 x+y^2+8 y=40 \\ & x^2-4 x+2^2+y^2+8 y+4^2=40+2^2+4^2 \\ & (x-2)^2+(y+4)^2=60 \end{aligned}$   Hence the centre of both circles is $(2,-4)$.   Let the radius of the required circle be $r$. $\therefore \quad r=\sqrt{(5-2)^2+(-2+4)^2}=\sqrt{13}$   Thus, the equation of the required circle is   $(x-2)^2+(y+4)^2=13$.

11. A circle passes through the points $P(-2,0), Q(3,5)$ and $R(6,2)$. Show that $P R$ is a diameter of the circle and find the equation of this circle.
    $\begin{aligned} &\text{Midpoint of } PR:\; \left(\frac{-2+6}{2}, \frac{0+2}{2}\right) = (2, 1) \\[4pt] &\text{Length of } PR:\; \sqrt{(6 - (-2))^2 + (2 - 0)^2} = \sqrt{64 + 4} = \sqrt{68} \\[4pt] &\text{Assume that $PR$ is a diameter.}\\[4pt] &\text{Radius } = \frac{\sqrt{68}}{2} = \sqrt{17} \\[4pt] &\text{Centre: } (2, 1),\quad \text{Radius: } \sqrt{17} \\[4pt] &\text{Circle: } (x - 2)^2 + (y - 1)^2 = 17 \\[4pt] &\text{Check if } Q(3,5)\text{ lies on it:} \\ &(3 - 2)^2 + (5 - 1)^2 = 1^2 + 4^2 = 1 + 16 = 17\;\checkmark \\ \therefore \quad& PR\text{ is a diameter and the equation is } (x - 2)^2 + (y - 1)^2 = 17 \end{aligned}$ Alternative Solution $\begin{aligned} &\text{Let the circle be } x^2 + y^2 + 2gx + 2fy + e = 0 \\[4pt] &\text{Substitute } P(-2,0):\; 4 - 4g + e = 0 \cdots\textbf{(1)} \\ &\text{Substitute } Q(3,5):\; 34 + 6g + 10f + e = 0 \;\cdots\textbf{(2)} \\ &\text{Substitute } R(6,2):\; 40 + 12g + 4f + e = 0 \;\cdots\textbf{(3)} \\[4pt] &\text{From (1): } e = 4g - 4,\; \text{ substitute into (2), (3)} \\[2pt] &(2):\; 34 + 6g + 10f + 4g - 4 = 0 \implies 10g + 10f = -30 \implies g + f = -3 \;\cdots\textbf{(4)} \\ &(3):\; 40 + 12g + 4f + 4g - 4 = 0 \implies 16g + 4f = -36 \;\cdots\textbf{(5)} \\[4pt] &\text{Solving equations (4) and (5): } \\ & g = -2,\; f = -1,\; e = -12 \\[4pt] &\text{Equation: } x^2 + y^2 - 4x - 2y - 12 = 0 \\ &(x - 2)^2 + (y - 1)^2 = 17 \\ &\text{Centre } (2,1),\; r = \sqrt{17} \\[4pt] &\text{Midpoint of } PR : \left(\frac{-2+6}{2}, \frac{0+2}{2}\right)= (2,1) \\ \therefore\quad& PR \text{ is a diameter. } \end{aligned}$

12. A circle passes through the points $(-1,-4)$ and $(3,2)$ and has radius $\sqrt{26}$. Find the two possible equations for this circle.
    Let the centre of the circle be $(h, k)$ Hence the equation of the circle is $(x-h)^2+(y-k)^2=2 6$. $(-1,-4)$ and $(3,2)$ lie on the circumference of the circle. $\begin{aligned} \therefore \quad & (-1-h)^2+(-4-k)^2=26 \\ & 1+2 h+h^2+16+8 k+k^2=26 \\ & h^2+k^2+2 h+8 k=9 \quad \ldots (1)\\ & (3-h)^2+(2-k)^2=26 \\ & 9-6 h+h^2+4-4 k+k^2=26 \\ & h^2+k^2-6 h-4 k=13 \quad \ldots (2) \end{aligned}$

13. A circle has radius 5 units and passes through the point $(2,-6)$. The $y$-axis is a tangent to the circle. Find the possible equations of the circle.
    Let the centre of the circle be $(h, k)$. Since the $y$-axis is tangent to the circle whose radius is 5, $h=5$. Hence, the equation of the circle is $(x-5)^2+(y-k)^2=25$. $(2,-6)$ lies on the circle $\begin{aligned} \therefore \quad & (2-5)^2+(-6-k)^2=25 \\ & (-6-k)^2=15 \\ \therefore \quad & -6-k=-4 \text{ or }-6-k=4 \\ \therefore \quad & k=-2 \text{ or } k=-10 \end{aligned}$ Thus, the possible equations of the circle are $(x-5)^2+(y+2)^2=25$ or $(x-5)^2+(y+10)^2=25$.

14. The equation of a circle is $(x-4)^2+(y+3)^2=41$. Show that the point $P(8,2)$ lies on the circle and find the equation of the tangent to the circle at the point $P$.
    Circle: $(x-4)^2+(y+3)^2=41$ When $x=8$ and $y=2$, $(8-4)^2+(2+3)^2=16+25=41$ Hence, $P(8,2)$ lies on the circle. Let the centre of the circle be $O$. Therefore, $O$ has coordinates $(4,-3)$. gradient of $O P=\frac{2-(-3)}{8-4}=\frac{5}{4}$ So, the gradient of tangent passing through $P$ is $-\frac{4}{5}$. The equation of tangent to the circle at $P$ is $\begin{aligned} & y-2=-\frac{4}{5}(x-8) \\ & 4 x+5 y=42 \end{aligned}$

15. Find the equation of the circle which passes through the points $(7,4)$ and $(0,5)$ and has its centre lying on the line $x+2 y=5$.
    $\begin{aligned} &\text{Let the centre of the circle be } (h,k)\\ & \text{ and radius be } r. \\ &\text{Since the circle passes through } (7,4) \text{ and } \\ &(0,5), \\ &(7-h)^2 + (4-k)^2 = (0-h)^2 + (5-k)^2 \\ &49 - 14h + h^2 + 16 - 8k + k^2 \\ &= h^2 + 25 - 10k + k^2 \\ &65 - 14h - 8k = 25 - 10k \\ &40 = 14h - 2k \\ &20 = 7h - k \cdots \textbf{\color{red}(1)} \\ &\text{The centre } (h,k) \text{ lies on the line } x+2y=5: \\ &h+2k=5 \cdots \textbf{\color{red}(2)} \end{aligned}$ $\begin{aligned} &h + 2(7h - 20) = 5 \quad ( \textbf{\small by equqtion (1)} )\\ &h + 14h - 40 = 5 \\ &15h = 45 \\ &h = 3 \\ &\text{Substitute } h=3 \text{ into } k = 7h - 20: \\ &k = 7(3) - 20 = 21 - 20 = 1 \\ &\text{The centre of the circle is } (3,1). \\ &r^2 = (0-3)^2 + (5-1)^2 \\ &r^2 = (-3)^2 + (4)^2 \\ &r^2 = 9 + 16 = 25 \\ &\text{The equation of the circle is }\\ & (x-3)^2 + (y-1)^2 = 25. \end{aligned}$

16. The points $A(-1,-4), B(10,7)$ and $C(13,4)$ are joined to form a triangle.

    1. Show that angle $A B C$ is a right angle.
    2. Find the equation of the circle that passes through the points $A, B$ and $C$.
    $\begin{aligned} \textbf{(a) } &\text{Slope of } AB = m_{AB} = \frac{7 - (-4)}{10 - (-1)} = \frac{11}{11} = 1 \\ &\text{Slope of } BC = m_{BC} = \frac{4 - 7}{13 - 10} = \frac{-3}{3} = -1 \\ &\text{Since } m_{AB} \times m_{BC} = 1 \times (-1) = -1, AB \perp BC, \\ \therefore\quad &\angle ABC \text{ is a right angle.} \\[10pt] \textbf{(b) } &\text{Since angle } ABC \text{ is a right angle, the hypotenuse } AC \text{ is the diameter of the circle.} \\ &\text{Midpoint of } AC = \left(\frac{-1+13}{2}, \frac{-4+4}{2}\right) = \left(\frac{12}{2}, \frac{0}{2}\right) = (6,0) \\ &\text{The centre of the circle is } (6,0). \\ & AC = \sqrt{(13 - (-1))^2 + (4 - (-4))^2}= \sqrt{260}= 2\sqrt{65} \\ &r =\frac{1}{2} AC= \frac{1}{2} \times 2\sqrt{65}= \sqrt{65} \\ &\text{The equation of the circle is } (x-6)^2 + y^2 = 65 \end{aligned}$

17. For each set of three points, find the equation of the circle that passes through them:
    (a) (4,3),(4,7),(6,9)

    (b) (0,5),(5,0),(0,-7)

    (c) (0,0),(4,2),(4,8)

    (d) (3,4),(1,-2),(6,3)

    (e) (-6,5),(-7,-2),(-3,6)
    $\begin{aligned} \textbf{(a) } &\text{Points: } (4,3),(4,7),(6,9) \\ &\text{Let the required circle be } x^2+y^2+2gx+2fy+e=0: \\ &(4,3): 16+9+8g+6f+e=0 \implies 8g+6f+e=-25\cdots \textbf{\color{red}(1)} \\ &(4,7): 16+49+8g+14f+e=0 \implies 8g+14f+e=-65\cdots \textbf{\color{red}(2)} \\ &(6,9): 36+81+12g+18f+e=0 \implies 12g+18f+e=-117 \cdots \textbf{\color{red}(3)} \\ &(2)-(1): 8f=-40 \implies f=-5 \\ \therefore\quad &8g-30+e=-25 \implies 8g+e=5 \cdots \textbf{\color{red}(4)} \\ &12g-90+e=-117 \implies 12g+e=-27 \cdots \textbf{\color{red}(5)} \\ &(5)-(4): 4g=-32 \implies g=-8 \\ \therefore\quad &8(-8)+e=5 \implies -64+e=5 \implies e=69 \\ \therefore \quad &\text{centre: } (-g,-f) = (8, 5)\\ &\text{radius} =\sqrt{g^2+f^2-e}=\sqrt{64+25-69}=\sqrt{20}\\ &\text{Equation of the circle: } (x-8)^2+(y-5)^2=20 \end{aligned}$ $\begin{aligned} \textbf{(b) } &\text{Points: } (0,5),(5,0),(0,-7) \\ &\text{Let the required circle be } x^2+y^2+2gx+2fy+e=0: \\ &(0,5): 0^2+5^2+2g(0)+2f(5)+e=0 \implies 25+10f+e=0 \cdots \textbf{\color{red}(1)} \\ &(5,0): 5^2+0^2+2g(5)+2f(0)+e=0 \implies 25+10g+e=0 \cdots \textbf{\color{red}(2)} \\ &(0,-7): 0^2+(-7)^2+2g(0)+2f(-7)+e=0 \implies 49-14f+e=0 \cdots \textbf{\color{red}(3)} \\ &(1)-(2): 10f-10g=0\implies f=g \\ &(1)-(3): -24 +24f =0 \implies f=1 \implies g=1 \\ \therefore\quad &25+ 10(1)+e=0 \implies e=-35 \\ \therefore \quad &\text{radius} =\sqrt{g^2+f^2-e}=\sqrt{1^2+1^2-(-35)}=\sqrt{1+1+35}=\sqrt{37}\\ &\text{centre: } (-g,-f) = (-1,-1). \\ &\text{Equation of the circle: } (x-(-1))^2+(y-(-1))^2=37 \\ &(x+1)^2+(y+1)^2=37 \end{aligned}$ $\begin{aligned} \textbf{(c) } &\text{Points: } (0,0),(4,2),(4,8) \\ &\text{Let the required circle be } x^2+y^2+2gx+2fy+e=0: \\ &(0,0): 0^2+0^2+2g(0)+2f(0)+e=0 \implies e=0 \\ &(4,2): 4^2+2^2+2g(4)+2f(2)+e=0 \implies 8g+4f+e=-20 \\ &\text{Since } e=0: 8g+4f=-20 \implies 2g+f=-5 \cdots \textbf{\color{red}(1)} \\ &(4,8): 4^2+8^2+2g(4)+2f(8)+e=0 \implies 8g+16f+e=-80 \\ &\text{Since } e=0: 8g+16f=-80 \implies g+2f=-10 \cdots \textbf{\color{red}(2)} \\ &(1) \times 2 \implies 4g+2f=-10 \cdots \textbf{\color{red}(3)} \\ &(3)-(2): 3g=0 \implies g=0 \\ \therefore\quad & 2(0)+f=-5 \implies f=-5 \\ \therefore \quad &\text{radius} =\sqrt{g^2+f^2-e}=\sqrt{0^2+(-5)^2-0}=\sqrt{25}=5 \\ &\text{centre: } (-g,-f) = (0,5). \\ &\text{Equation of the circle: } (x-0)^2+(y-5)^2=25 \\ &x^2+(y-5)^2=25 \end{aligned}$ $\begin{aligned} \textbf{(d) } &\text{Points: } (3,4),(1,-2),(6,3) \\ &\text{Let the required circle be } x^2+y^2+2gx+2fy+e=0: \\ &(3,4): 3^2+4^2+2g(3)+2f(4)+e=0 \implies 6g+8f+e=-25 \cdots \textbf{\color{red}(1)} \\ &(1,-2): 1^2+(-2)^2+2g(1)+2f(-2)+e=0 \implies 2g-4f+e=-5 \cdots \textbf{\color{red}(2)} \\ &(6,3): 6^2+3^2+2g(6)+2f(3)+e=0 \implies 12g+6f+e=-45 \cdots \textbf{\color{red}(3)} \\ &(1)-(2): 4g+12f=-20 \implies g+3f=-5 \cdots \textbf{\color{red}(4)} \\ &(3)-(1): 6g-2f=-20 \implies 3g-f=-10 \cdots \textbf{\color{red}(5)} \\ &(5)\times 3\implies 9g-3f=-30 \cdots \textbf{\color{red}(6)} \\ &(4)+(6): 10g=-35 \implies g=-7/2 \\ \therefore\quad &\text{From (4), } (-7/2)+3f=-5 \implies f=-1/2 \\ \therefore\quad &2(-7/2)-4(-1/2)+e=-5 \implies e=0 \\ \therefore \quad &\text{radius} =\sqrt{g^2+f^2-e}=\sqrt{\bigg(-\frac{7}{2}\bigg)^2+\bigg(\frac{7}{2}\bigg)^2-0}=\sqrt{\frac{25}{2}} \\ &\text{centre: } (-g,-f) = \bigg(\frac{7}{2},\frac{1}{2}\bigg). \\ &\text{Equation of the circle: } \bigg(x-\frac{7}{2}\bigg)^2+\bigg(y-\frac{1}{2}\bigg)^2=\frac{25}{2} \end{aligned}$ $\begin{aligned} \textbf{(e) } &\text{Points: } (-6,5),(-7,-2),(-3,6) \\ &\text{Let the required circle be } x^2+y^2+2gx+2fy+e=0: \\ &(-6,5): (-6)^2+5^2+2g(-6)+2f(5)+e=0 \implies -12g+10f+e=-61 \cdots \textbf{\color{red}(1)} \\ &(-7,-2): (-7)^2+(-2)^2+2g(-7)+2f(-2)+e=0 \implies -14g-4f+e=-53 \cdots \textbf{\color{red}(2)} \\ &(-3,6): (-3)^2+6^2+2g(-3)+2f(6)+e=0 \implies -6g+12f+e=-45 \cdots \textbf{\color{red}(3)} \\ &(1)-(2): 2g+14f=-8 \implies g+7f=-4 \cdots \textbf{\color{red}(4)} \\ &(3)-(1): 6g+2f=16 \implies 3g+f=8 \cdots \textbf{\color{red}(5)} \\ &(5)\times 7: 21g+7f=56 \cdots \textbf{\color{red}(6)} \\ &(6)-(4): 20g=60 \implies g=3 \\ \therefore\quad &\text{From (5), } 3(3)+f=8 \implies 9+f=8 \implies f=-1 \\ &\text{Substitute } g=3, f=-1 \text{ into (1):} \\ &-12(3)+10(-1)+e=-61 \implies e=-15 \\ \therefore \quad &\text{radius} =\sqrt{g^2+f^2-e}=\sqrt{3^2+(-1)^2-(-15)}=5 \\ &\text{centre: } (-g,-f) = (-3,1). \\ &\text{Equation of the circle: } (x-(-3))^2+(y-1)^2=25 \\ &(x+3)^2+(y-1)^2=25 \end{aligned}$

21. Use each of the following methods to show that the four points $A(-4,-2)$, $B(-2,2), C(5,1)$ and $D(4,-6)$ lie on a circle.

    1. Methods
       1. Use the property that the perpendicular bisector of two chords on a circle intersect at the centre of the circle to find the equation of the circle passing through three of the points and then show that the fourth point lies on this circle.
       2. Use three of the points to write down three simultaneous equations. Solve the equations and then show that the fourth point lies on this circle.
       3. Show that opposite angles in the quadrilateral $A B C D$ add up to $180^{\circ}$ which means that $A B C D$ is a cyclic quadrilateral. Hence the four points lie on a circle.
       4. Show that angle $B A D=90^{\circ}$. Hence $B D$ must be a diameter of the circle (angles in a semi-circle) and the centre of the circle is the midpoint of $B D$. Then show that the fourth point, $C$, lies on this circle.
    2. Find out about Ptolemy's theorem and use it to show that the points $A, B, C$ and $D$ lie on a circle.
    (a) Method 1 Midpoint of $A B=\left(\frac{-4-2}{2}, \frac{-2+2}{2}\right)=(-3,0)$ Midpoint of $A D=\left(\frac{-4+4}{2}, \frac{-2-6}{2}\right)=(0,-4)$ gradient of $A B=\frac{2+2}{-2+4}=2$ gradient of $A D=\frac{-6+2}{4+4}=-\frac{1}{2}$ Let the perpendicular bisector of $A B$ and $A D$ be $l_1$ and $l_2$ respectively. gradient of $l_1=-\frac{1}{2}$ gradient of $l_2=2$ Equation of $l_1\implies y-0=-\frac{1}{2}(x+3) \implies y=-\frac{1}{2}(x+3) \ldots(1)$ Equation of $l_2\implies y=2 x-4 \ldots(2)$ Solving equations (1) and (2) $\begin{aligned} & -\frac{1}{2}(x+3)=2 x-4 \\ & 5 x=5 \implies x=1 \\ \therefore\quad & y=2-4=-2 \end{aligned}$ Hence, the centre of the circle passing through $A, B$ and $D$ is $(1,-2)$. radius of circle $=\sqrt{(1+4)^2+(-2+2)^2}=5$ Therefore, the equation of the circle passing through $A, B$ and $D$ is $(x-1)^2+(y+2)^2=25$ When $x=5$ and $y=1$, $(5-1)^2+(1+2)^2=4^2+3^2=25$ $\therefore\quad A, B, C$ and $D$ all lie on a circle. (a) Method 2 Let the equation of the circle be $x^2+y^2+2 g x+2 f y+e=0$ At $A(-4,-2)$, $\begin{aligned} & 16+4-8 g-4 f+e=0 \\ & -8 g-4 f+e=-20 \ldots(1) \end{aligned}$ At $B(-2,2)$, $\begin{aligned} & 4+4-4 g+4 f+e=0 \\ & -4 g+4 f+e=-8 \ldots(2) \end{aligned}$ At $D(4,-6)$, $\begin{aligned} & 16+36+8 g-12 f+e=0 \\ & 8 g-12 f+e=-52 \ldots(3) \end{aligned}$ Equation (1) + Equation (2) $\begin{aligned} & -12 g+2 e=-28 \\ & e=6 g-14 \ldots(4) \end{aligned}$ Substitute $e=6 g-14$ in equations (2) and (3), $\begin{aligned} & -4 g+4 f+6 g-14=-8 \\ & g=3-2 f \ldots(5) \\ & 8 g-12 f+6 g-14=-52 \\ & 7 g-6 f=-19 \ldots(6) \\ & 21-14 f-6 f=-19 \implies f=2 \end{aligned}$ $\begin{aligned} & g=3-2(2)=-1 \\ & e=6(-1)-14=-20 . \end{aligned}$ The equation of circle passing through $A, B$ and $D$ is $x^2+y^2-2 x+4 y-20=0$ When $x=5$ and $y=1$, $25+1-10+4-20=0$ Hence, the points $A, B, C$ and $D$ all lie on a circle. (a) Method 3 $\begin{aligned} & m_{A B}=\frac{2+2}{-2+4}=2 \\[2mm] & m_{B C}=\frac{1-2}{5+2}=-\frac{1}{7} \\[2mm] & m_{C D}=\frac{-6-1}{4-5}=7 \\[2mm] & m_{A D}=\frac{-6+2}{4+4}=-\frac{1}{2} \\[2mm] \therefore \quad & m_{A B} \cdot m_{A D}=2\left(-\frac{1}{2}\right)=-1 \\ \therefore \quad & A B \perp A D \implies \angle B A D=90^{\circ} \\ \therefore \quad & m_{B C} \cdot m_{C D}=-\frac{1}{7}(7)=-1 \\ & B C \perp C D \implies \angle B C D=90^{\circ} \\ \therefore \quad & \angle B A D+\angle B C D=180^{\circ}\implies A B C D \text{ is a cycic quadrilateral.} \end{aligned}$ (a) Method 4 $\begin{aligned} & m_{A B}=\frac{2+2}{-2+4}=2 \\ & m_{A D}=\frac{-6+2}{4+4}=-\frac{1}{2} \\ & m_{A B} \cdot m_{A D}=2\left(-\frac{1}{2}\right)=-1 \\ \therefore \quad & A B \perp A D \implies \angle B A D=90^{\circ} \end{aligned}$ $\therefore \quad B D$ is the diameter of circle passing through $A, B$ and $D$. $\begin{aligned} \therefore \text{ centre } & =\left(\frac{4-2}{2}, \frac{-6+2}{2}\right)=(1,-2) \\ \text{ radius } & =\sqrt{(4-1)^2+(-6+2)^2}=5 \end{aligned}$ $\therefore\ $ The equation of the circle through $A, B$ and $D$ is $(x-1)^2+(y+2)^2=25$ When $x=5$ and $y=1$, $(5-1)^2+(1+2)^2=16+9=25 \text{ . }$ $\therefore\ A, B, C$ and $D$ all lie on a circle. (b) Ptolemy's theorem If a quadrilateral is cyclic then the product of the lengths of its diagonals is equal to the sum of the products of the lengths of the pairs of opposite sides. (ref: wikipedia) $\begin{aligned} & A(-4,-2), B(-2,2), C(5,1), D(4,-6) \\ & A B=\sqrt{(-2+4)^2+(2+2)^2}=\sqrt{20}=2 \sqrt{5} \\ & B C=\sqrt{(5+2)^2+(1-2)^2}=\sqrt{50}=5 \sqrt{2} \\ & C D=\sqrt{(4-5)^2+(-6-1)^2}=\sqrt{50}=5 \sqrt{2} \\ & A D=\sqrt{(4+4)^2+(-6+2)^2}=\sqrt{80}=4 \sqrt{5} \\ & B D=\sqrt{(4+2)^2+(-6-2)^2}=10 \\ & A C=\sqrt{(5+4)^2+(1+2)^2}=\sqrt{90}=3 \sqrt{10}\\ \therefore\quad & B D \cdot A C=30 \sqrt{10} \\ & A B \cdot C D=10 \sqrt{10} \\ & B C \cdot A D=20 \sqrt{10} \\ \therefore\quad & A B \cdot C D+B C \cdot A D=30 \sqrt{10} \\ \therefore\quad & B D \cdot A C=A B \cdot C D+B C \cdot A D\implies A B C D \text{ is a cycic quadrilateral.} \end{aligned}$

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