Circle 3 Question 5

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

(2019 Main,10 Jan I)

(a) 5

(b) $2 \sqrt{5}$

(c) $\sqrt{57}$

(d) 4

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Answer:

Correct Answer: 5. (a)

Solution:

  1. Equation of tangent to the circle $x^{2}+y^{2}+4 x-6 y-12=0$ at $(1,-1)$ is given by $x x _1+y y _1+2\left(x+x _1\right)-3\left(y+y _1\right)-12=0$, where $x _1=1$ and $y _1=-1$

$\Rightarrow x-y+2(x+1)-3(y-1)-12=0$

$\Rightarrow 3 x-4 y-7=0$

This will also a tangent to the required circle.

Now, equation of family of circles touching the line $3 x-4 y-7=0$ at point $(1,-1)$ is given by

$(x-1)^{2}+(y+1)^{2}+\lambda(3 x-4 y-7)=0$

So, the equation of required circle will be $(x-1)^{2}+(y+1)^{2}+\lambda(3 x-4 y-7)=0$, for some $\lambda \in R$ …(i)

$\because$ The required circle passes through $(4,0)$

$\therefore(4-1)^{2}+(0+1)^{2}+\lambda(3 \times 4-4 \times 0-7)=0$

$\Rightarrow 9+1+\lambda(5)=0 \Rightarrow \lambda=-2$

Substituting $\lambda=-2$ in Eq. (i), we get

$(x-1)^{2}+(y+1)^{2}-2(3 x-4 y-7)=0$

$\Rightarrow \quad x^{2}+y^{2}-8 x+10 y+16=0$

On comparing it with

$$ \begin{aligned} & x^{2}+y^{2}+2 g x+2 f y+c=0, \text { we get } \\ & g=-4, f=5, c=16 \\ \therefore \quad & \text { Radius }=\sqrt{g^{2}+f^{2}-c}=\sqrt{16+25-16}=5 \end{aligned} $$



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