Two Circles Graph
Exercise 7.2
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Find the points of intersection of the line and the circle:(a) $y=4$ and $x^2+y^2-4 x-6 y+8=0$(b) $y=x-1$ and $x^2+y^2+5 x-y-6=0$(c) $y=2 x+3$ and $x^2+y^2-4 x+6 y-12=0$(d) $y=2 x+1$ and $x^2+y^2+5 x-3 y-12=0$
$\begin{aligned} \textbf{(a) } &\text{Line: } y=4 \\ &\text{Circle: } x^2+y^2-4 x-6 y+8=0 \\ &\text{Substitute } y=4 \text{ into the circle equation:} \\ &x^2+(4)^2-4x-6(4)+8=0 \\ &x^2+16-4x-24+8=0 \\ &x^2-4x=0 \\ &x(x-4)=0 \\ &x=0 \text{ or } x=4 \\ &\text{Points of intersection: } (0,4) \text{ and } (4,4) \end{aligned}$
$\begin{aligned} \textbf{(b) } &\text{Line: } y=x-1 \\ &\text{Circle: } x^2+y^2+5 x-y-6=0 \\ &\text{Substitute } y=x-1 \text{ into the circle equation:} \\ &x^2+(x-1)^2+5x-(x-1)-6=0 \\ &x^2+(x^2-2x+1)+5x-x+1-6=0 \\ &2x^2+2x-4=0 \\ &x^2+x-2=0 \\ &(x+2)(x-1)=0 \\ &x=-2 \text{ or } x=1 \\ &\text{If } x=-2, y=-2-1=-3 \implies (-2,-3) \\ &\text{If } x=1, y=1-1=0 \implies (1,0) \\ &\text{Points of intersection: } (-2,-3) \text{ and } (1,0) \end{aligned}$
$\begin{aligned} \textbf{(c) } &\text{Line: } y=2x+3 \\ &\text{Circle: } x^2+y^2-4 x+6 y-12=0 \\ &\text{Substitute } y=2x+3 \text{ into the circle equation:} \\ &x^2+(2x+3)^2-4x+6(2x+3)-12=0 \\ &x^2+(4x^2+12x+9)-4x+12x+18-12=0 \\ &5x^2+20x+15=0 \\ &x^2+4x+3=0 \\ &(x+1)(x+3)=0 \\ &x=-1 \text{ or } x=-3 \\ &\text{If } x=-1, y=2(-1)+3=1 \implies (-1,1) \\ &\text{If } x=-3, y=2(-3)+3=-3 \implies (-3,-3) \\ &\text{Points of intersection: } (-1,1) \text{ and } (-3,-3) \end{aligned}$
$\begin{aligned} \textbf{(d) } &\text{Line: } y=2x+1 \\ &\text{Circle: } x^2+y^2+5 x-3 y-12=0 \\ &\text{Substitute } y=2x+1 \text{ into the circle equation:} \\ &x^2+(2x+1)^2+5x-3(2x+1)-12=0 \\ &x^2+(4x^2+4x+1)+5x-6x-3-12=0 \\ &5x^2+3x-14=0 \\ &(5x-7)(x+2)=0 \\ & x=7/5 \text{ or } x=-2 \\ &\text{If } x=7/5, y=2(7/5)+1 = 19/5 \\ &\text{If } x=-2, y=2(-2)+1 = -3 \\ &\text{Points of intersection: } (7/5,19/5) \text{ and } (-2,-3) \end{aligned}$ -
By considering the radii of each circle and the distance between their centres, determine whether each pair of circles intersect, touch or do not intersect.(a) $ x^2+y^2=25$ and $x^2+y^2-15 x+8 y-75=0$(b) $ x^2+y^2=16$ and $x^2+y^2+3 x-5 y-6=0$(c) $x^2+y^2=25$ and $x^2+y^2-24 x-18 y+125=0$(d) $(x-8)^2+y^2=4$ and $(x-2)^2+(y+5)^2=1$(e) $ x^2+(y-3)^2=9$ and $(x-8)^2+(y-5)^2=144$
$\begin{aligned} \textbf{(a) } &C_1: x^2+y^2=25\\ & \text{Centre } (0,0),\quad r_1=5 \\ &C_2: x^2+y^2-15 x+8 y-75=0 \\ & \text{Centre } \left(\frac{15}{2}, -4\right),\quad r_2=\sqrt{\left(\frac{15}{2}\right)^2+(-4)^2-(-75)}=\sqrt{\frac{589}{4}} \\ &d = \sqrt{\left(\frac{15}{2}-0\right)^2+(-4-0)^2} =8.5 \\ &r_1+r_2 = 5+\frac{\sqrt{589}}{2} \approx 17.11 \\ &|r_1-r_2| = \left|5-\frac{\sqrt{589}}{2}\right| \approx 7.11 \\ &\text{Since } |r_1-r_2| < d < r_1+r_2, \text{ the circles intersect.} \\[10pt] \end{aligned}$
$\begin{aligned} \textbf{(b) } &C_1: x^2+y^2=16 \\ & \text{Centre } (0,0), \quad r_1=4 \\ &C_2: x^2+y^2+3 x-5 y-6=0 \\ & \text{Centre } \left(-\frac{3}{2}, \frac{5}{2}\right),\quad r_2=\sqrt{\left(-\frac{3}{2}\right)^2+\left(\frac{5}{2}\right)^2-(-6)}= \sqrt{\frac{29}{2}} \\ &d = \sqrt{\left(-\frac{3}{2}-0\right)^2+\left(\frac{5}{2}-0\right)^2} =\sqrt{\frac{17}{2}} \approx 2.92 \\ &r_1+r_2 = 4+\sqrt{\frac{29}{2}} \approx 7.81 \\ &|r_1-r_2| = \bigg|4-\sqrt{\frac{29}{2}}\bigg| \approx 0.19 \\ &\text{Since } |r_1-r_2| < d < r_1+r_2, \text{ the circles intersect.} \end{aligned}$
$\begin{aligned} \textbf{(c) } &C_1: x^2+y^2=25\\ & \text{Centre } (0,0), \quad r_1=5 \\ &C_2: x^2+y^2-24 x-18 y+125=0 \\ &\text{Centre } (12, 9),\quad r_2=\sqrt{12^2+9^2-125}=10 \\ &d = \sqrt{(12-0)^2+(9-0)^2} =15 \\ &r_1+r_2 = 5+10=15 \\ &\text{Since } d = r_1+r_2, \text{ the circles touch externally.} \end{aligned}$
$\begin{aligned} \textbf{(d) }&C_1: (x-8)^2+y^2=4\\ & \text{Centre } (8,0),\quad r_1=2 \\ &C_2: (x-2)^2+(y+5)^2=1 \\ & \text{Centre } (2,-5),\quad r_2=1 \\ &d = \sqrt{(8-2)^2+(0-(-5))^2} = \sqrt{61} \\ &r_1+r_2 = 2+1=3 \\ &\text{Since } d > r_1+r_2 , \text{ the circles do not intersect.}\\ \end{aligned}$
$\begin{aligned} \textbf{(e) } &C_1: x^2+(y-3)^2=9\\ &\text{Centre } (0,3),\quad r_1=3 \\ &C_2: (x-8)^2+(y-5)^2=144 \\ & \text{Centre } (8,5),\quad r_2=12 \\ &d = \sqrt{(8-0)^2+(5-3)^2} = \sqrt{8^2+2^2}=\sqrt{64+4}=\sqrt{68} \\ &r_1+r_2 = 3+12=15 \\ &|r_1-r_2| = |3-12|=9 \\ &\text{Since } d < |r_1-r_2| , \text{ the smaller circle is inside the larger circle and they do not intersect.} \end{aligned}$ -
The line $y=x-3$ intersects the circle $(x-3)^2+(y+2)^2=20$ at two points $P$ and $Q$. Find the length of $P Q$.Line: $y=x-3$
Circle: $(x-3)^2+(y+2)^2=20$
At the points of intersection of the line and circle,
$\begin{aligned} & (x-3)^2+(x-3+2)^2=20 \\ & x^2-6 x+9+x^2-2 x+1-20=0 \\ & 2 x^2-8 x-10=0 \\ & x^2-4 x-5=0 \\ & (x+1)(x-5)=0 \\ & x=-1 \text{ or } x=5 \\ & x=-1 \implies y=-4 \\ & x=5 \implies y=2 \end{aligned}$
Hence, $P$ and $Q$ have coordinates $(-1,-4)$ and $(5,2)$.
$P Q=\sqrt{(5+1)^2+(2+4)^2}=6 \sqrt{2}$ -
Explain why the line $4 y=x+26$ is a tangent to the circle $x^2+y^2+10 x-2 y+9=0$ and find the coordinates of the point where the line touches the curve.Line: $4 y=x+26 \implies x=4 y-26$
Circle: $x^2+y^2+10 x-2 y+9=0$
At the point of intersection of the line and circle,
$\begin{aligned} & (4 y-26)^2+y^2+10(4 y-26)-2 y+9=0 \\ & 16 y^2-208 y+676+y^2+40 y-260-2 y+9=0 \\ & 17 y^2-170 y+425=0 \end{aligned}$$\begin{aligned} & y^2-10 y+25=0 \\ & (y-5)^2=0\implies y=5\\ \therefore\quad& x=-6 \end{aligned}$
Hence the line intersects the circle at only one point $(-6,5)$.
Thus, the line is tangent to the circle. -
Explain why the circles $(x+7)^2+(y+2)^2=1$ and $(x+7)^2+(y+5)^2=16$ touch at one point. Determine whether they touch internally or externally and find the coordinates of the point at which they touch.
$\begin{aligned} &\text{Circle 1: } (x+7)^2 + (y+2)^2 = 1\implies \text{centre } C_1 = (-7, -2),\quad r_1 = 1 \\ &\text{Circle 2: } (x+7)^2 + (y+5)^2 = 16 \implies \text{centre } C_2 = (-7, -5),\quad r_2 = 4 \\ &\text{Distance between centres: } d = \sqrt{(-7 + 7)^2 + (-2 + 5)^2} = \sqrt{0 + 9} = 3 \\ &\text{Difference of radii: } |r_1 - r_2| = |1 - 4| = 3 \\ &\text{Since } d = |r_1 - r_2|, \text{the circles touch internally}. \\ &\text{At the point of contact, } \\ &(y+5)^2 -(y+2)^2 = 15\\ &y^2+10y+25-y^2-4y-4=15\\ &y=-1\\ &\text{Since the two circles lie on the same vertical line $x=-7$, the point of contact is } (-7, -1) \end{aligned}$ -
The circles $(x-10)^2+(y-5)^2=25$ and $(x-20)^2+(y-10)^2=100$ intersect at the points $P$ and $Q$.
- Find the length of the line $P Q$.
- Find the equation of the common chord $P Q$.
Method 1
Circle 1: $\quad(x-10)^2+(y-5)^2=25$
centre: $(10,5)$, radius $=5$
Circle 2: $\quad(x-20)^2+(y-10)^2=100$
centre: $(20,10)$, radius $=10$
Let the centre of the two circles be $M(10,5)$ and $N(20,10)$ and the radii of the circles be $r_1=5 $ and $r_2=10 $.
$M N=\sqrt{(20-10)^2+(10-5)^2}=5 \sqrt{5} $
Let the point of intersection of $P Q$ and $MN$ be $R$.
Let $MR=k$.
$\therefore \quad R N=5 \sqrt{5}-k$
$\begin{aligned} \therefore \quad & P R^2=5^2-k^2 \text{ or } \\ & P R^2=10^2-(5 \sqrt{5}-k)^2 \\ \therefore \quad & 5^2-k^2=10^2-(5 \sqrt{5}-k)^2 \\ & 25-k^2=100-125+10 \sqrt{5} k-k^2 \\ \end{aligned}$$\begin{aligned} & 10 \sqrt{5} k=50 \implies k=\sqrt{5} \\ \therefore \quad& P R^2=25-5=20 \implies P R=2 \sqrt{5} \\ \therefore\quad & P Q=2 P R=4 \sqrt{5} \end{aligned}$
gradient of $M N=\frac{10-5}{20-10}=\frac{1}{2}$
$\therefore\quad $ gradient of $P Q=-2\quad (\because \ P Q \perp \mathrm{MN})$
$\frac{M R}{R N}=\frac{\sqrt{5}}{4 \sqrt{5}}=\frac{1}{4}$
By section formula,
$ R=\left(\frac{4 (10)+20}{5}, \frac{4 (5)+10}{5}\right)=(12,6)$
Equation of $PQ$ is
$y-6 =-2(x-12) \implies y =30-2 x .$
Method 2
$\begin{aligned} & \text{Circle} 1: (x-10)^2+(y-5)^2=25\\ & x^2-20 x+100+y^2-10 y+25-25=0 \\ & x^2+y^2-20 x-10 y+100=0 \ldots(1) \end{aligned}$
$\begin{aligned} & \text{Circle} 2:(x-20)^2+(y-10)^2=100\\ & x^2-40 x+400+y^2-20 y+100-100=0 \\ & x^2+y^2-40 x-20 y+400=0 \ldots(2) \end{aligned}$
At the points of intersection two circles,
$\begin{aligned} & x^2+y^2-20 x-10 y+100=x^2+y^2-40 x-20 y+400 \\ & 10 y=300-20 x \implies y=30-2 x \ldots(3) \end{aligned}$
Substituting $y=30-2 x$ in equation ( 1 )
$\begin{aligned} & x^2+(30-2 x)^2-20 x-10(30-2 x)+100=0 \\ & x^2+300-120 x+4 x^2-20 x-300+20 x+100= 0\\ & 5 x^2-120 x+700=0 \\ & x^2-24 x+140=0 \\ & (x-10)(x-14)=0 \\ & x=10 \text{ or } x=14 \\ & x=10 \implies y=10 \\ & x=14 \implies y=2 . \\ & P Q=\sqrt{(14-10)^2+(2-10)^2}=4 \sqrt{5} \end{aligned}$
Equation of $P Q: \quad y=30-2 x$. -
Show that the line $y=2 x-1$ and the circle $x^2+y^2+15 x-11 y+9=0$ do not intersect.$\begin{aligned} &\text{Substitute } y = 2x - 1 \text{ into the circle equation:} \\ &x^2 + (2x - 1)^2 + 15x -11(2x -1) + 9 = 0 \\ &x^2 + 4x^2 - 4x + 1 + 15x -22x + 11 + 9 = 0 \\ & 5x^2 -11x + 21 = 0 \\ &\text{Discriminant } = (-11)^2 - 4 (5) ( 21) = -299 <0\\ \therefore \quad & \text{The equation has no real solution}\\ \therefore \quad &\text{The line and circle do not intersect.} \end{aligned}$
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The equation of a circle is $x^2+y^2-6 x-8 y=0$.
- Show that the point $P(7,7)$ lies on this circle.
- Find the equation of the tangent to the circle at the point $P$.
$\begin{aligned} \textbf{(a) }&x^2+y^2-6 x-8 y=0 \\ &\text{When $x=7$ and $y=7$, }\\ &7^2+7^2-6 (7)-8 (7)=0= 25 \quad \text{(Correction: $49+49-42-56=0$)} \\ \therefore \quad &\text{the point } P(7,7) \text{ lies on the circle}\\ &x^2+y^2-6 x-8 y=0 \\ &x^2-6 x+9+y^2-8 y+16=25 \\ &(x - 3)^2 + (y - 4)^2 = 25 \\ &\text{Centre is } (3, 4),\quad \text{radius } r = 5\\ &\text{Slope of radius } CP = \frac{7 - 4}{7 - 3} = \frac{3}{4} \\ &\text{Slope of tangent } = -\frac{4}{3} \\ &\text{Using point-slope form} \\ &y - 7 = -\frac{4}{3}(x - 7) \\ &3y - 21 = -4x + 28 \\ &4x + 3y = 49 \\ &\text{Equation of tangent is } 4x + 3y = 49 \end{aligned}$ -
The line $2 x+y=6$ intersects the circle $x^2+y^2-12 x-8 y+27=0$ at the points $A$ and $B$.
- Find the coordinates of the points $A$ and $B$.
- Find the equation of the perpendicular bisector of $A B$.
- The perpendicular bisector of $A B$ intersects the circle at the points $P$ and $Q$. Find the exact coordinates of $P$ and $Q$.
- Find the exact area of quadrilateral $A P B Q$.
Line: $2 x+y=6 \implies y=6-2 x$
Circle: $x^2+y^2-12 x-8 y+27=0$
(a) At the point of intersection of the line and circle,
$\begin{aligned} & x^2+(6-2 x)^2-12 x-8(6-2 x)+27=0 \\ & x^2+36-24 x+4 x^2-12 x-48+16 x+27=0 \\ & 5 x^2-20 x+15=0 \\ & x^2-4 x+3=0 \\ & (x-1)(x-3)=0 \\ & x=1, x=3 \\ & x=1 \implies y=4 \\ & x=3 \implies y=0 \end{aligned}$
Thus, $A$ and $B$ have coordinates $(1,4)$ and $(3,0)$ respectively.
Let $M$ be the midpoint of $A B$ and $l$ be the perpendicular bisector of $A B$.
$\therefore\ M=\left(\frac{1+3}{2}, \frac{4+0}{2}\right)=(2,2)$
gradient of $A B=\frac{4-0}{1-3}=-2$
gradient of $l=\frac{1}{2}$
Equation of $l$ : $y-2=\frac{1}{2}(x-2) \implies 2 y-x=2 \text{ (or) } x=2 y-2$
At the point of intersection of $l$ and circle,
$\begin{aligned} & (2 y-2)^2+y^2-12(2 y-2)-8 y+27=0 \\ & 4 y^2-8 y+4+y^2-24 y+24-8 y+27=0 \\ & 5 y^2-40 y+55=0\\ & y^2-8 y+11=0 \\ & y^2-8 y+16=5 \\ & (y-4)^2=5 \end{aligned}$
$\begin{aligned} & y=4 \pm \sqrt{5} \\ & y=4-\sqrt{5}, x=6-2 \sqrt{5} \\ & y=4+\sqrt{5}, x=6+2 \sqrt{5} \end{aligned}$
P and Q have coordinates $(6-2 \sqrt{5}, 4-\sqrt{5}) \text{ and }(6+2 \sqrt{5}, 4+\sqrt{5}) $.
The circle $x^2+y^2-12 x-8 y+27=0$ has radius $v=\sqrt{36+16-27}=5$.
Since $P Q$ is the diameter of the circle, $P Q=10$.
$A B=\sqrt{(3-1)^2+(0-4)^2}=2 \sqrt{5}$
area of $A P B Q=\frac{1}{2} \times A B \times P Q=\frac{1}{2} \times 2 \sqrt{5} \times 10=10 \sqrt{5}$ -
Show that these pairs of circles touch each other and find the coordinates of the point where they touch. In each case, sketch the circles to show how they touch.
- $x^2+y^2-8 x-4 y+19=0$ and $(x-5)^2+(y-2)^2=4$
- $x^2+y^2+6 x-4 y+3=0$ and $x^2+y^2-12 x+2 y-3=0$
$\begin{aligned} \textbf{(a) } & \text{Circle 1: } x^2 + y^2 - 8x - 4y + 19 = 0 \\ &x^2 - 8x+4^2 + y^2- 4y+2^2 = -19+4^2+2^2 \\ & (x - 4)^2 + (y - 2)^2 = 1 \\ & \text{centre } (4,2),\quad r_1= 1 \\ &\text{Circle 2: } (x-5)^2+(y-2)^2=4\\ & \text{Circle 2: centre } (5,2), \quad r_2= 2\\ & \text{Distance between centres: } d = \sqrt{(5 - 4)^2 + (2 - 2)^2} = 1 \\ & r_1+r_2= 1 + 2 = 3,\quad |r_1-r_2| = |2 - 1 |= 1 \\ & \text{Since $d= |r_1-r_2|$ they touch internally}\\ & \text{At the point of contact} \\ &(x-5)^2 -(x - 4)^2=3\\ &9-2x=3\implies x=3\\ &\text{Since the two circles lies on the same horizontal line $y=2$, the point of contact is } (3, 2)\\[10pt] \textbf{(b) } & \text{Circle 1: }x^2 + y^2 + 6x - 4y + 3 = 0 \implies (x + 3)^2 + (y - 2)^2 = 10 \\ & \text{Centre } (-3,2), \quad r_1= \sqrt{10} \\ & \text{Circle 2: }x^2 + y^2 - 12x + 2y - 3 = 0 \implies (x - 6)^2 + (y + 1)^2 = 40 \\ & \text{Centre } (6,-1), \quad r_2=\sqrt{40}=2 \sqrt{10}\\ & \text{Distance between centres:} \\ & d=\sqrt{(6 + 3)^2 + (-1 - 2)^2} = \sqrt{90}=3\sqrt{10} \\ \end{aligned}$
$\begin{aligned} &r_1+r_2= \sqrt{10}+2 \sqrt{10}=3\sqrt{10}\\ & \text{So, circles touch externally} \\ & \text{At the point of contact, } \\ & 18 x-6 y+6=0 \\ & y=3 x+1\\ &\text{Substitute } y=3 x+1 \text{ in Circle } 1 \text{ equation }\\ & x^2+(3 x+1)^2+6 x-4(3 x+1)+3=0 \\ & x^2+9 x^2+6 x+1+6 x-12 x-4+3=0 \\ & 10 x^2=0\implies x=0 \implies y=1\\ \therefore \quad&\text{ the point of contact of the two circles is } (0, 1) \end{aligned}$ -
Find the set of values of $m$ for which the line $y=m x+3$ intersects the circle $x^2+y^2-10 x-8 y+28=0$ at two distinct points.
Line: $y=m x+3$
Circle: $x^2+y^2-10 x-8 y+28=0$
At the point of intersection the line and circle,
$\begin{aligned} & x^2+(m x+3)^2-10 x-8(m x+3)+28=0 \\ & x^2+m^2 x^2+6 m x+9-10 x-8 m x-2 y+28=0 \\ & \left(m^2+1\right) x^2-2(m+5) x+13=0 \end{aligned}$
Since the line intersects the circle at two distinct points, discriminant $>0$
$\begin{aligned} & 4(m+5)^2-4\left(m^2+1\right)(13)>0 \\ & -6 m^2+5 m+6>0 \implies 6 m^2-5 m+6<0 \implies (3 m+2)(2 m-3)<0 \implies -\frac{2}{3} -
Two circles have the following properties:
- the $x$-axis is a common tangent to the circles
- the point $(14,2)$ lies on both circles
- the centre of each circle lies on the line $x+2 y=28$
$\begin{aligned} &\text{Let the center of the first circle be } (x, y).\\ &\text{Since the $x$-axis is tangent, radius } = y.\\ &\text{Point } (14,2) \text{ lies on the circle: } \sqrt{(14 - x)^2 + (2 - y)^2} = y \\ &(14 - x)^2 + (2 - y)^2 = y^2 \\ &196 - 28x + x^2 + 4 - 4y + y^2 = y^2 \\ &x^2 - 28x + 200 - 4y = 0 \quad\cdots(1) \\ &\text{Centre lies on } x + 2y = 28 \implies x = 28 - 2y \quad\cdots(2)\\ &\text{Substitute (2) into (1):} \\ &(28 - 2y)^2 - 28(28 - 2y) + 200 - 4y = 0 \\ & 784 - 112y + 4y^2 - 784 + 56y + 200 - 4y = 0 \\ & 4y^2 - 60y + 200 = 0 \implies y^2 - 15y + 50 = 0 \\ &y = 5 \text{ or } 10 \\ &\text{If } y = 5, x = 28 - 2(5) = 18; \quad \text{If } y = 10, x = 28 - 2(10) = 8 \\ \therefore\quad & \text{Centres are } (18,5) \text{ and } (8,10)\\ &\text{The equations of the circles are } (x-18)^2+(y-5)^2=25 \text{ and }(x-8)^2+(y-10)^2=100. \end{aligned}$
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