A circle touches the y-axis at the point (0, | vedclass

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

Question

(C) Let the center of the circle be $O' = (h, k)$.Since the circle touches the $y$-axis at $(0, 4)$,the $y$-coordinate of the center is $k = 4$ and th

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Let the center of the circle be $O' = (h, k)$.Since the circle touches the $y$-axis at $(0, 4)$,the $y$-coordinate of the center is $k = 4$ and the radius $r = |h|$.Thus,the equation of the circle is $(x - h)^2 + (y - 4)^2 = h^2$.This circle cuts the $x$-axis $(y = 0)$ at points where $(x - h)^2 + (0 - 4)^2 = h^2$,which simplifies to $(x - h)^2 + 16 = h^2$,or $(x - h)^2 = h^2 - 16$.So,$x - h = \pm \sqrt{h^2 - 16}$,which gives $x = h \pm \sqrt{h^2 - 16}$.The length of the chord on the $x$-axis is the difference between these $x$-values: $(h + \sqrt{h^2 - 16}) - (h - \sqrt{h^2 - 16}) = 2\sqrt{h^2 - 16}$.Given the chord length is $6$,we have $2\sqrt{h^2 - 16} = 6$,so $\sqrt{h^2 - 16} = 3$.Squaring both sides,$h^2 - 16 = 9$,which gives $h^2 = 25$,so $h = 5$.Therefore,the radius $r = |h| = 5$.

$A$ circle touches the $y$-axis at the point $(0, 4)$ and cuts the $x$-axis in a chord of length $6$ units. The radius of the circle is

Similar Questions

The equation of the circle passing through the points $(0, 0)$,$(0, b)$,and $(a, b)$ is

The equation of the circle concentric with the circle $x^2+y^2-6x-4y-12=0$ and touching the $Y$-axis is:

If $x = \frac{2at}{1+t^2}$ and $y = \frac{a(1-t^2)}{1+t^2}$,where $t$ is a parameter,then $a$ represents:

$A$ circle touches the axes at the points $(3, 0)$ and $(0, -3)$. The centre of the circle is

The equation of the circle concentric with the circle $x^2 + y^2 + 8x + 10y - 7 = 0$ and passing through the centre of the circle $x^2 + y^2 - 4x - 6y = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module