Circle Ques 94
- The locus of the centre of a circle, which touches externally the circle $x^{2}+y^{2}-6 x-6 y+14=0$ and also touches the $Y$-axis, is given by the equation $(1993,1 M)$
(a) $x^{2}-6 x-10 y+14=0$
(b) $x^{2}-10 x-6 y+14=0$
(c) $y^{2}-6 x-10 y+14=0$
(d) $y^{2}-10 x-6 y+14=0$
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Answer:
Correct Answer: 94.(d)
Solution:
Formula:
Common Tangents of Two Circles
- Let $(h, k)$ be the centre of the circle which touches the circle $x^{2}+y^{2}-6 x-6 y+14=0$ and $Y$-axis.
The centre of given circle is $(3,3)$ and radius is $\sqrt{3^{2}+3^{2}-14}=\sqrt{9+9-14}=2$
Since, the circle touches $Y$-axis, the distance from its centre to $Y$-axis must be equal to its radius, therefore its radius is $h$. Again, the circles meet externally, therefore the distance between two centres $=$ sum of the radii of the two circles.
Hence,
$$ (h-3)^{2}+(k-3)^{2}=(2+h)^{2} $$
$$ h^{2}+9-6 h+k^{2}+9-6 k=4+h^{2}+4 h $$
i.e.
$$ k^{2}-10 h-6 k+14=0 $$
Thus, the locus of $(h, k)$ is
$$ y^{2}-10 x-6 y+14=0 $$
BETA
