The centres of those circles which touch the | vedclass

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

Question

(B) The given circle is $x^{2} + y^{2} - 8x - 8y - 4 = 0$.The center of this circle is $(4, 4)$ and its radius $r = \sqrt{4^{2} + 4^{2} - (-4)} = \sqr

Vedclass · Accepted Answer

a parabola

Vedclass · Answer

(B) The given circle is $x^{2} + y^{2} - 8x - 8y - 4 = 0$.The center of this circle is $(4, 4)$ and its radius $r = \sqrt{4^{2} + 4^{2} - (-4)} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6$.Let the center of the required circle be $(h, k)$ and its radius be $R$. Since the circle touches the $x$-axis,its radius $R = |k|$. Assuming the circle is above the $x$-axis,$R = k$.Since the circle touches the given circle externally,the distance between their centers is equal to the sum of their radii:$\sqrt{(h - 4)^{2} + (k - 4)^{2}} = 6 + k$.Squaring both sides:$(h - 4)^{2} + (k - 4)^{2} = (6 + k)^{2}$.Expanding the terms:$h^{2} - 8h + 16 + k^{2} - 8k + 16 = 36 + k^{2} + 12k$.Simplifying the equation:$h^{2} - 8h - 20k - 4 = 0$.Replacing $(h, k)$ with $(x, y)$,we get the locus:$x^{2} - 8x - 20y - 4 = 0$.This is the equation of a parabola.

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:

Similar Questions

Let $C$ be the centre and $A$ be one end of a diameter of the circle $x^2+y^2-2x-4y-20=0$. If $P$ is a point such that $A$ divides $CP$ in the ratio $2:3$,then the locus of $P$ is

The locus of a point which is at a distance of $4$ units from $(3, -2)$ in the $xy$-plane 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:

$A$ point moves in such a way that its distance from $(1, -2)$ is always twice its distance from $(-3, 5)$. The locus of the point is:

The locus of the centers of the circles that are passing through the intersection of the circles $x^2+y^2=1$ and $x^2+y^2-2x+y=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module