Question

The centre of those circles which touches the circle, ${x^2} + {y^2} - 8x - 8y - 4 = 0$ externally and also touch the x-axis, lie on:
(A) a circle
(B) an ellipse which is not a circle
(C) a hyperbola
(D) a parabola

Answer 1

Firstly, Compare the given circle${x^2} + {y^2} - 8x - 8y - 4 = 0$ with the general equation a circle, i.e., ${x^2} + {y^2} + 2gx + 2fy + c = 0$ and find out the centre of circle $\left( { - g, - f} \right)$ and radius of circle $\sqrt {{g^2} + {f^2} - c}$. As the two circles touch externally, so equate the distance between their centers and sum of their radii to find out the required result.

Complete step-by-step answer:
Given circle is ${C_1} \Rightarrow$ ${x^2} + {y^2} - 8x - 8y - 4 = 0$
Compare this given equation with the general equation a circle i.e., ${x^2} + {y^2} + 2gx + 2fy + c = 0$, we get-
$2g = - 8 \Rightarrow g = - 4$
$2f = - 8 \Rightarrow f = - 4$
$c = - 4$
Centre of circle$= \left( { - g, - f} \right) = \left( {4,4} \right)$
Radius of circle$= \sqrt {{g^2} + {f^2} - c} = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2} - \left( { - 4} \right)} = \sqrt {16 + 16 + 4} = \sqrt {36} = 6$
Let a circle ${C_2} \Rightarrow {\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$ touches the given circle ${C_1}$.
Center of circle ${C_2}$ is $\left( {h,k} \right)$ and radius is $r$.
We know that when two circles touch externally, then the distance between their centers is equal to the sum of their radii.
Therefore, distance between the centers $\left( {h,k} \right)$ and $\left( {4,4} \right)$= sum of radii= $r + 6$
$\Rightarrow$ $\sqrt {{{\left( {h - 4} \right)}^2} + {{\left( {k - 4} \right)}^2}} = r + 6$
On squaring both sides, we get-
${\left( {h - 4} \right)^2} + {\left( {k - 4} \right)^2} = {\left( {r + 6} \right)^2}$ …. (1)
Since ${C_2}$ touch the x-axis , therefore put $r = k$ in above equation (1)
${\left( {h - 4} \right)^2} + {\left( {k - 4} \right)^2} = {\left( {k + 6} \right)^2}$
$\Rightarrow {\left( {h - 4} \right)^2} + {k^2} + 16 - 8k = {k^2} + 36 + 12k$
On simplifying the terms, we get-
$\Rightarrow {\left( {h - 4} \right)^2} = 20k + 20$
$\Rightarrow {\left( {h - 4} \right)^2} = 4\left( {5k + 5} \right)$
Now, replacing $\left( {h,k} \right)$ by $\left( {x,y} \right)$-
$\Rightarrow {\left( {x - 4} \right)^2} = 4\left( {5y + 5} \right)$
This equation is in the form of parabola. Hence, the locus is parabola.

So, the correct answer is “Option D”.

Note: When two circles touch externally, then the distance between their centers is equal to the sum of their radii. When a circle touches the x-axis, its radius becomes $k$ and if the circle touches the y-axis, its radius becomes $h$.