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

Open in App

Solution

Construction $x^{2} + y^{2} + 8 x - 8 y - 4 = 0$
$\to$ radius $= \sqrt{4^{2} + 4^{2} + 4} = \sqrt{36} = 6$
center $(- 4, 4)$
construct a line parallel to $x = 0$ at a distance of $6$ units from it as shown
$\frac{C P}{P M} = \frac{r + 6}{r + 6} = 1$
$\Rightarrow \frac{C P}{P M} = 1$
$\Rightarrow$ P is a point whose distance from fixed point 'C' and its distance from fixed line $^{'} L_{1}^{'}$ are equal
From locus definition P(center) lies on parabola.

Suggest Corrections

Similar questions

x^{2} + y^{2} - 8 x - 8 y - 4 = 0

x

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

x^{2} + y^{2} = 1

y -

x^{2} + y^{2} = 1

x^{2} + y^{2} - 6 x - 6 y + 14 = 0

y

Join BYJU'S Learning Program