Question

The circles $x^2+y^2-4x-81=0,x^2+y^2+24x-81=0$ intersect each other at points $A$ and $B$. A line through point $A$ meet one circle at $P$ and a parallel line through $B$ meet the other circle at $Q$. Then the locus of the mid point of $PQ$ is

• A. $(x+5{)}^2+(y+0{)}^2=25$
• B. $(x-5{)}^2+(y-0{)}^2=25$
• C. $x^2+y^2+10x=0$
• D. $x^2+y^2-10x=0$

Answer 1

The correct option is A $(x+5{)}^2+(y+0{)}^2=25$

Given equation of circles

$(x+12{)}^2+y^2=(15{)}^2......\left(1\right)C_1(x-2{)}^2+y^2=85.........(2)C_2$

Point of intersection of two circles

Subtracting $\left(2\right)$ from $\left(1\right)$, we get

$(x+12{)}^2-(x-2{)}^2=140(2x+10)\left(14\right)=140x=0,y=\pm 9$

Let $P(x_1,y_1)$ on circle $C_1$ and $Q(x_2,y_2)$ be the point on circle $C_2$

$PA\parallel BQ$

Equation of line passing through $P$ and $A$

$(y-9)=m(x-0)\Rightarrow y=mx+9$

Equation of line passing through $B$ and $Q$

$(y+9)=m(x-0)\Rightarrow y=mx-9$

$P$ is on circle $C_1$ and $Q$ is on circle $C_2$

Intersection of line $(y=mx+9)$ with circle $C_1$ at point $P$

$(x_1+12{)}^2+(m{x}_1+9{)}^2=\left(15{)}^2\right(1+m^2)x^2+(24-18m)x_1=0$

$x_1=0,\frac{18m-24}{1+m^2}$ gives $y_1=9,\frac{9m^2-24m-9}{1+m^2}$

Intersection of line $y=mx-9$ with circle $C_1$ at $Q$

$(x_1-5{)}^2+y^2=85(x_1-2{)}^2+(m{x}_1-9{)}^2=85(1+m^2)x_1+(18m-4)x_2=0$

$x_2=0,\frac{4-18m}{1+m^2}$ gives $y_2=-9,\frac{-9m^2+4m+9}{1+m^2}$

$(h,k)$ is the midpoint of $P$ and $Q$

$h=\frac{x_1+x_2}{2}$ and $k=\frac{y_1+y_2}{2}$

Substituting the value of $x_1,x_2,y_1,y_2$, we get

$h=\frac{-10}{1+m^2}$ and $k=\frac{-10m}{1+m^2}$

$h=\frac{-10}{1+m^2}k=\frac{100m^2}{(1+m^2{)}^2}......\left(3\right)$

$h+5=\frac{5+5m^2-10}{1+m^2}=\frac{5(m^2-1)}{1+m^2}$

Squaring both sides $(h+5{)}^2=25\frac{(m^2-1{)}^2}{(1+m^2{)}^2}.....(4)$

Adding $\left(3\right)$ and $\left(4\right)$

$(h+5{)}^2+k^2=25\frac{(m^2-1{)}^2+100m^2}{(1+m^2{)}^2}$

$(h+5{)}^2+k^2=25\frac{[m^4+1-2m^2+4m^2]}{(1+m^2{)}^2}$

$(h+5{)}^2+k^2=25\frac{(1+m^2{)}^2}{(1+m^2{)}^2}(h+5{)}^2+k^2=25$

$(x+5{)}^2+y^2=25$