Circle 4 Question 12
12. Two circles $x^{2}+y^{2}=6$ and $x^{2}+y^{2}-6 x+8=0$ are given. Then the equation of the circle through their points of intersection and the point $(1,1)$ is
$(1980,1 M)$
(a) $x^{2}+y^{2}-6 x+4=0$
(b) $x^{2}+y^{2}-3 x+1=0$
(c) $x^{2}+y^{2}-4 y+2=0$
(d) None of these
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Answer:
Correct Answer: 12. (b)
Solution:
- The required equation of circle is, $S _1+\lambda\left(S _2-S _1\right)=0$.
$$ \therefore \quad\left(x^{2}+y^{2}-6\right)+\lambda(-6 x+14)=0 $$
Also, passing through $(1,1)$.
$$ \begin{aligned} \Rightarrow & -4+\lambda(8) & =0 \\ \Rightarrow & \lambda & =\frac{1}{2} \end{aligned} $$
$\therefore$ Required equation of circle is
$$ \text { or } \quad \begin{aligned} x^{2}+y^{2}-6-3 x+7 & =0 \\ x^{2}+y^{2}-3 x+1 & =0 \end{aligned} $$
