Circle 2 Question 9
10. If the circle $x^{2}+y^{2}+2 x+2 k y+6=0$ and $x^{2}+y^{2}+2 k y+k=0$ intersect orthogonally, then $k$ is
(a) 2 or $-3 / 2$
(b) -2 or $-3 / 2$
(c) 2 or $3 / 2$
(d) -2 or $3 / 2$
$(2000,2 M)$
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Answer:
Correct Answer: 10. (a)
Solution:
- Since, the given circles intersect orthogonally.
$ \begin{aligned} & \therefore & 2(1)(0)+2(k) & (k)=6+k \\ & & & {\left[\because 2 g _1 g _2+2 f _1 f _2=c _1+c _2\right] } \\ & & & \\ & & 2 k^{2}-k-6=0 \Rightarrow & k=-\frac{3}{2}, 2 \end{aligned} $
