Circle Ques 112
- The centres of those circles which touch the circle, $x^{2}+y^{2}-8 x-8 y-4=0$, externally and also touch the $X$-axis, lie on
(2016 Main)
(a) a circle
(b) an ellipse which is not a circle
(c) a hyperbola
(d) a parabola
Show Answer
Answer:
Correct Answer: 112.(d)
Solution:
Formula:
Common Tangents of Two Circles
- Given equation of circle is
$x^{2}+y^{2}-8 x-8 y-4=0$, whose centre is $C(4,4)$ and radius $=\sqrt{4^{2}+4^{2}+4}=\sqrt{36}=6$
Let the centre of required circle be $C _1(x, y)$. Now, as it touch the $X$-axis, therefore its radius $=|y|$. Also, it touch the circle
$$ \begin{array}{lc} & x^{2}+y^{2}-8 x-8 y-4=0 \text {, therefore } C C _1=6+|y| \\ \Rightarrow & \sqrt{(x-4)^{2}+(y-4)^{2}}=6+|y| \\ \Rightarrow & x^{2}+16-8 x+y^{2}+16-8 y=36+y^{2}+12|y| \\ \Rightarrow & x^{2}-8 x-8 y+32=36+12|y| \\ \Rightarrow & x^{2}-8 x-8 y-4=12|y| \end{array} $$
Case I If $y>0$, then we have
$$ \begin{aligned} & & x^{2}-8 x-8 y-4 & =12 y \\ \Rightarrow & & x^{2}-8 x-20 y-4 & =0 \\ \Rightarrow & & x^{2}-8 x-4 & =20 y \end{aligned} $$
$\Rightarrow \quad(x-4)^{2}-20=20 y$
$\Rightarrow \quad(x-4)^{2}=20(y+1)$ which is a parabola.
Case II If $y<0$, then we have
$$ \begin{array}{lc} & x^{2}-8 x-8 y-4=-12 y \\ \Rightarrow & x^{2}-8 x-8 y-4+12 y=0 \\ \Rightarrow & x^{2}-8 x+4 y-4=0 \\ \Rightarrow & x^{2}-8 x-4=-4 y \\ \Rightarrow & (x-4)^{2}=20-4 y \\ \Rightarrow & (x-4)^{2}=-4(y-5) \text {, which is again a parabola. } \end{array} $$
BETA
