Question

If a circle $C$ passing through $(4,0)$ touches the circle $x^2+y^2+4x-6y-12=0$ externally at a point $(1,-1)$, then the radius of the circle $C$ is:

• A. $\sqrt{57}$
• B. $2\sqrt{5}$
• C. $4$
• D. $5$

Answer 1

Answer: (D)

$5$

Letthe equation of circle be,

$(x-h{)}^2+(y-k{)}^2=r^2$

$(h,k)$ is centre and $r$ is radius

and, $(y-h{)}^2+k^2=r^2$

$(l-h{)}^2+(l+k{)}^2=r^2$

Now, the centre of circle $C$ will be lie on the line passing through point of contact and centre of the circle $x^2+y^2+4x-by-12=0$

which has an equation

$\frac{y+1}{x-1}=\frac{+3+1}{-2-1}$

$\Rightarrow -3y-3=4x-4$

$\Rightarrow 4x+3y=1$

$(h,k)$ satisfies this equation so,

$4h+3k=1$

so, $3k=1-4h$

$k=\frac{1-4h}{3}$

so,

$(4-h{)}^2+{\left(\frac{1-4h}{3}\right)}^2=r^2$

$\Rightarrow 16+h^2-8h+\frac{1+16h^2-8h}{9}=r^2$

$\Rightarrow \frac{25}{a}{h}^2-\frac{80h}{9}+\frac{145}{9}=r^2.......\left(1\right)$

and, $(1-h{)}^2+{(1+\frac{1-4h}{3})}^2=r^2$

$\Rightarrow 1+h^2-2h+{\left(\frac{4-4h}{3}\right)}^2=r^2$

$\Rightarrow 1+h^2-2h+\frac{16+16h^2-32h}{9}=r^2$

$\Rightarrow \frac{25h^2}{9}-\frac{50h}{9}+\frac{25}{9}=r^2......\left(2\right)$

solving $\left(1\right)$ & $\left(2\right)$ we get,

$\frac{25h^2}{9}-\frac{50h}{9}+\frac{25}{9}=\frac{25h^2}{9}-\frac{80h}{9}+\frac{145}{9}$

$\Rightarrow \frac{30h}{9}=\frac{120}{9}$

$\Rightarrow h=4$

so, $k=\frac{1-16}{3}=-5$

so, $r=\left[\right(4-4{)}^2+(-5{)}^2{]}^{1/2}$

$\Rightarrow r=5$