Circle Ques 67
- 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: 67.(a)
Solution:
Formula:
- 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
$$ \begin{aligned} & x^{2}+y^{2}+2 g x+2 f y+c=0, \text { we get } \\ & g=-4, f=5, c=16 \\ \therefore \quad & \text { Radius }=\sqrt{g^{2}+f^{2}-c}=\sqrt{16+25-16}=5 \end{aligned} $$
BETA
