The locus of the centres of the circles,which | vedclass

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

Question

(D) Let the centre of the circle be $(h, k)$ and its radius be $r$. Since the circle touches the $y$-axis,the radius $r = |h| = h$ (as it lies in the

Vedclass · Accepted Answer

$y = \sqrt{1 + 2x}, x \ge 0$

Vedclass · Answer

(D) Let the centre of the circle be $(h, k)$ and its radius be $r$. Since the circle touches the $y$-axis,the radius $r = |h| = h$ (as it lies in the first quadrant,$h > 0$). Since the circle touches $x^2 + y^2 = 1$ externally,the distance between the centres is equal to the sum of the radii: $\sqrt{h^2 + k^2} = r + 1 = h + 1$. Squaring both sides,we get: $h^2 + k^2 = (h + 1)^2$ $h^2 + k^2 = h^2 + 2h + 1$ $k^2 = 2h + 1$. Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = 2x + 1$,or $y = \sqrt{2x + 1}$ for $x \ge 0$.

The locus of the centres of the circles,which touch the circle $x^2 + y^2 = 1$ externally,also touch the $y$-axis and lie in the first quadrant is

Similar Questions

$A$ circle touches the $x$-axis and also touches the circle with centre at $(0, 3)$ and radius $2$. The locus of the centre of the circle is

The radius of the circle with minimum area that touches the curve $y = 4 - x^2$ and the lines $y = |x|$ is:

Let a variable line passing through the centre of the circle $x^2+y^2-16x-4y=0$ meet the positive coordinate axes at points $A$ and $B$. Then the minimum value of $OA+OB$,where $O$ is the origin,is equal to

The locus of the mid-points of chords of the circle $x^{2}+y^{2}=1$,which subtend a right angle at the origin,is

The locus of a point $P(x, y)$ satisfying the equation $\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=4$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module