A circle touches the x-axis and cuts off a ch | vedclass

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

Question

(D) Let the circle be $(x - h)^2 + (y - k)^2 = r^2$,where $(h, k)$ is the centre and $r$ is the radius.Since the circle touches the $x$-axis,the radiu

Vedclass · Accepted Answer

$A$ hyperbola

Vedclass · Answer

(D) Let the circle be $(x - h)^2 + (y - k)^2 = r^2$,where $(h, k)$ is the centre and $r$ is the radius.Since the circle touches the $x$-axis,the radius $r = |k|$.Thus,the equation of the circle is $(x - h)^2 + (y - k)^2 = k^2$,which simplifies to $x^2 - 2hx + h^2 + y^2 - 2ky = 0$.The circle cuts a chord of length $2l$ from the $y$-axis (where $x = 0$).Substituting $x = 0$ into the equation,we get $y^2 - 2ky + h^2 = 0$.The length of the intercept on the $y$-axis is given by $2\sqrt{k^2 - h^2} = 2l$.Squaring both sides,we get $k^2 - h^2 = l^2$.Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 - x^2 = l^2$,which represents a hyperbola.

$A$ circle touches the $x$-axis and cuts off a chord of length $2l$ from the $y$-axis. The locus of the centre of the circle is

Similar Questions

$A$ line through $(0,0)$ cuts the circle $x^2 + y^2 - 2ax = 0$ at points $A$ and $B$. The locus of the centre of the circle drawn on $AB$ as a diameter is:

The locus of the point whose ratio of distance from the origin to its distance from $(-2, -3)$ is $5: 7$,is given by:

$A$ circle is drawn to cut a chord of length $2a$ units along the $X$-axis and to touch the $Y$-axis. The locus of the centre of the circle is

The locus of the point which is equidistant from the point $(1,1)$ and the line $x+y+1=0$ is

If the locus of the midpoints of the chords of the circle $x^2+y^2=25$,which subtend a right angle at the origin,is given by $\frac{x^2}{a^2}+\frac{y^2}{a^2}=1$,then $|a|=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module