Circle 5 Question 3

3. 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

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Answer:

Correct Answer: 3. (d)

Solution:

  1. 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

$$ \begin{array}{lc} & x^{2}-8 x-8 y-4=-12 y \\ \Rightarrow & x^{2}-8 x-8 y-4+12 y=0 \\ \Rightarrow & x^{2}-8 x+4 y-4=0 \\ \Rightarrow & x^{2}-8 x-4=-4 y \\ \Rightarrow & (x-4)^{2}=20-4 y \\ \Rightarrow & (x-4)^{2}=-4(y-5) \text {, which is again a parabola. } \end{array} $$



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