The locus of the centres of all circles which | vedclass

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Question

(A) Let the centre of the circle be $(h, k)$ and its radius be $r$.Since the circle touches the line $x=2a$,the distance from the centre $(h, k)$ to t

Vedclass · Accepted Answer

$y^2+4ax-5a^2=0$

Vedclass · Answer

(A) Let the centre of the circle be $(h, k)$ and its radius be $r$.Since the circle touches the line $x=2a$,the distance from the centre $(h, k)$ to the line $x-2a=0$ is equal to the radius $r$.Thus,$r = |h-2a|$.Squaring both sides,$r^2 = (h-2a)^2 = h^2 - 4ah + 4a^2$.The circle is $x^2+y^2-2hx-2ky+c=0$,where $c = h^2+k^2-r^2$.Substituting $r^2$,we get $c = h^2+k^2-(h^2-4ah+4a^2) = k^2+4ah-4a^2$.The circle cuts $x^2+y^2=a^2$ orthogonally,so $2g_1g_2 + 2f_1f_2 = c_1+c_2$.Here,$g_1 = -h, f_1 = -k, c_1 = c$ and $g_2 = 0, f_2 = 0, c_2 = -a^2$.Thus,$2(-h)(0) + 2(-k)(0) = c - a^2$.This implies $c = a^2$.Equating the two expressions for $c$: $k^2+4ah-4a^2 = a^2$.$k^2+4ah = 5a^2$.Replacing $(h, k)$ with $(x, y)$,the locus is $y^2+4ax=5a^2$,or $y^2+4ax-5a^2=0$.

The locus of the centres of all circles which touch the line $x=2a$ and cut the circle $x^2+y^2=a^2$ orthogonally is:

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