Let the circles pass through the point (-1, 1 | vedclass

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

Question

(B) The equation of a circle with center $(h, k)$ tangent to the $x$-axis is $(x - h)^2 + (y - k)^2 = k^2$.Since the circle passes through the point $

Vedclass · Accepted Answer

$k \geq 1/2$

Vedclass · Answer

(B) The equation of a circle with center $(h, k)$ tangent to the $x$-axis is $(x - h)^2 + (y - k)^2 = k^2$.Since the circle passes through the point $(-1, 1)$,we have $(-1 - h)^2 + (1 - k)^2 = k^2$.Expanding this,we get $(h + 1)^2 + 1 - 2k + k^2 = k^2$,which simplifies to $h^2 + 2h + 2 - 2k = 0$.This is a quadratic equation in $h$: $h^2 + 2h + (2 - 2k) = 0$.For $h$ to be a real value,the discriminant $D$ must be greater than or equal to $0$.$D = b^2 - 4ac = (2)^2 - 4(1)(2 - 2k) \geq 0$.$4 - 8 + 8k \geq 0$.$8k - 4 \geq 0$.$8k \geq 4$.$k \geq 1/2$.

Let the circles pass through the point $(-1, 1)$ and be tangent to the $x$-axis. If $(h, k)$ are the coordinates of the center of the circle,then the set of values of $k$ is given by which interval?

Similar Questions

The equation of the circle which touches both the axes and whose radius is $a$ is

The circle $x^2 + y^2 - 3x - 4y + 2 = 0$ cuts the $x$-axis at

If $z=x+iy$,then the centre of the circle $\left|\frac{z-3}{z-2i}\right|=2$ is

Let $S$ be a circle concentric with the circle $3x^2+3y^2+x+y-1=0$. If the length of the tangent drawn from a point $(2,-2)$ to the given circle is the radius of the circle $S$,then the power of the point $(2,1)$ with respect to the circle $S$ is

$A$ circle touches both axes and its center lies in the fourth quadrant. If its radius is $1$,then its equation is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module