The locus of the centres of all circles which | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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:

Similar Questions

The locus of the centre of the circles which touch both the circles $x^{2}+y^{2}=a^{2}$ and $x^{2}+y^{2}=4ax$ externally is

If a system of circles passes through $(2,3)$ and cuts the circle $x^2+y^2=12$ orthogonally,then the equation of the locus of the centres of that system of circles is:

Let the locus of the midpoints of the chords of the circle $x^2+(y-1)^2=1$ drawn from the origin intersect the line $x+y=1$ at $P$ and $Q$. Then,the length of $PQ$ is:

The locus of the centre of a circle touching both the coordinate axes is

If $P(x_1, y_1)$ is a point such that the lengths of the tangents from it to the circles $x^2+y^2-4x-6y-12=0$ and $x^2+y^2+6x+18y+26=0$ are in the ratio $2:3$,then the locus of $P$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module