Question

the locus of the center of a circle which touches externally the circle x^2+y^2-6x-6y+14=0 and also touches the y-axis is given by the equation

Answer 1

hope it will help you!!!!!!!!

Answer 2

Concept

A locus may be a set of points that meet a given condition. The definition of a circle locus of points a given distance from a given point during a two-dimensional plane. The given distance is that the radius and therefore the given point is that the center of the circle.

Given

The equation of the circle is $x^2+y^2-6x-6y+14=0$.

Find

we have to search out the locus of the middle.

Solution

Let $(h, k)$ be the centre of the circle which touches the circle $x^2+y^2-6x-6y+14=0$ and$x$-axis.

The centre of the given circle is $(3,3)$ and therefore the radius is

$\sqrt{32} + 3^2 - 14 = 9 + 9 - 14 = 2$

Since the circle touches the $y$-axis, the space from its centre to the $y$-axis must be adequate to its radius, therefore its radius is h. The circles touch one another externally, hence, the gap between two centres equal the sum of the radii of the $2$ circles.

$\begin{aligned}(h - 3)^2 + (k -3)^2& = (2 + h)^2\\ h^2 + 9 -6h + k^2 + 9 -6k& = 4 + h^2 + 4h\\ k^2-10h-6k+14&=0\end{aligned}$

The locus of $(h, k)$is$y^2 -10x - 6y + 14 = 0$.

Hence, the locus of the middle of the circle is $y^2 -10x - 6y + 14 = 0$.

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