Question

Lines $5x+12y-10=0$ and $5x-12y-40=0$ touch a circle $C_1$ of diameter $6$. If the centre of $C_1$ lies in the first quadrant, find the equation of the circle $C_2$ which is concentric with $C_1$ and cuts intercepts of length $8$ on these lines.

• A. $x^2+y^2-10x-4y+4=0$
• B. $x^2+y^2+10x-4y+4=0$
• C. $x^2+y^2-10x+4y-6=0$
• D. $x^2+y^2-10x+4y+6=0$

Answer 1

Answer: (A)

$x^2+y^2-10x-4y+4=0$

Since the lines $5x+12y-10=0$ and $5x-12y-40=0$ touches the circle $C_1$,

its center lies on one of the angle bisector of the given lines, i.e. on the lines

$\frac{5x+12y-10}{\sqrt{25+144}}=\pm \frac{5x-12y-40}{\sqrt{25+144}}\Rightarrow 5x+12y-10=\pm 5x-12y-40$

That is $x=5$ or $y=-\frac{5}{4}$

Since the center of $C_1$ lies in the first quadrant.

Let its coordinates be $(5,t)$ where $t>0$

As $5x+12y-10=0$ touches $C_1$ we have

$\left|\frac{5\times 5+12t-10}{\sqrt{25+144}}\right|=3$ (radius of $C_1$)

$\Rightarrow 15+12t=\pm 3\times 13=\pm 39$

That is $t=2$ or $t=-\frac{9}{2}$

Now, we must take $t=2$, so that the center of $C_1$ is $(5,2)$ in the first quadrant

If $r$ is the radius of the circle $C_2$, we have

$r^2=3^2+4^2\Rightarrow r=5$

Hence the equation of $C_2$ is

${(x-5)}^2+{(y-2)}^2=5^2\Rightarrow x^2+y^2-10x-4y+4=0$