The centres of those circles which touch the circle, x^2 + y^2 - 8x - 8y - 4 = 0, external…

Question

The centres of those circles which touch the circle, ${x^2} + {y^2} - 8x - 8y - 4 = 0$, externally and also touch the $x$-axis, lie on :

Accepted Answer

For the given circle,

center : $(4,4)$

radius $=6$

$$6 + k = \sqrt {{{\left( {h - 4} \right)}^2} + {{\left( {k - 4} \right)}^2}} $$

$${\left( {h - 4} \right)^2} = 20k + 20$$

$$\therefore$$ locus of $(h, k)$ is

$${\left( {h - 4} \right)^2} = 20\left( {y + 1} \right),$$

which is parabola.

Solutions · 1