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

$A$ circle $x^2 + y^2 + 2gx + 2fy + c = 0$ passing through $(4, -2)$ is concentric to the circle $x^2 + y^2 - 2x + 4y + 20 = 0$. Then the value of $c$ is:

Find the equation of the circle passing through the points $(2, 3)$ and $(-1, 1)$ and whose centre is on the line $x - 3y - 11 = 0$.

Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ be two points such that their abscissae $x_1$ and $x_2$ are the roots of the equation $x^2 + 2x - 3 = 0$,while the ordinates $y_1$ and $y_2$ are the roots of the equation $y^2 + 4y - 12 = 0$. The centre of the circle with $PQ$ as diameter is

The circle $x^2 + y^2 + 6y = 0$ touches which of the following?

The centres of the circles $x^2 + y^2 = 1$,$x^2 + y^2 + 6x - 2y = 1$ and $x^2 + y^2 - 12x + 4y = 1$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module