The circle possessing y-axis as its tangent a | vedclass

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

Question

(D) Let $(h, k)$ be the center of the circle. Since the circle touches the $y$-axis at $(0,2)$,the radius $r = |h|$ and the $y$-coordinate of the cent

Vedclass · Accepted Answer

$(-4,0)$

Vedclass · Answer

(D) Let $(h, k)$ be the center of the circle. Since the circle touches the $y$-axis at $(0,2)$,the radius $r = |h|$ and the $y$-coordinate of the center $k = 2$. The equation of the circle is $(x-h)^2 + (y-2)^2 = h^2$. Since it passes through $(-1, 0)$,we substitute these coordinates: $(-1-h)^2 + (0-2)^2 = h^2$ $1 + 2h + h^2 + 4 = h^2$ $2h = -5 \Rightarrow h = -\frac{5}{2}$. The equation of the circle is $\left(x + \frac{5}{2}\right)^2 + (y-2)^2 = \left(\frac{5}{2}\right)^2$. Checking option $(d)$ $(-4, 0)$: $\left(-4 + \frac{5}{2}\right)^2 + (0-2)^2 = \left(-\frac{3}{2}\right)^2 + 4 = \frac{9}{4} + 4 = \frac{25}{4} = \left(\frac{5}{2}\right)^2$. Thus,the circle passes through $(-4, 0)$.

The circle possessing $y$-axis as its tangent at $(0,2)$ and passing through $(-1,0)$,also passes through

Similar Questions

The equation $ax^2 + 2y^2 + 2bxy + 2x - y + c = 0$ represents a circle passing through the origin,if:

The equation of a circle which passes through the points $(2, 3)$ and $(4, 5)$ and whose center lies on the straight line $y - 4x + 3 = 0$ is

The equation of the circle in the first quadrant touching each coordinate axis at a distance of one unit from the origin is

If the endpoints of a diameter of a circle are $(0, 1)$ and $(1, 1)$,then its equation is . . . .

$A$ circle cuts off positive intercepts $5$ and $6$ on the $x$ and $y$ axes respectively,and passes through the origin. Then the equation of the circle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module