Find the equation of the circle passing throu | vedclass

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

Question

(A) Step-$1$: Find the radius $r$ of the circle.The radius is the distance between the center $(2, -1)$ and the point $(3, 6)$ on the circle.Using the

Vedclass · Accepted Answer

$x^2 + y^2 - 4x + 2y - 45 = 0$

Vedclass · Answer

(A) Step-$1$: Find the radius $r$ of the circle.The radius is the distance between the center $(2, -1)$ and the point $(3, 6)$ on the circle.Using the distance formula $r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$r = \sqrt{(3 - 2)^2 + (6 - (-1))^2}$$r = \sqrt{(1)^2 + (7)^2} = \sqrt{1 + 49} = \sqrt{50}$.Step-$2$: Use the standard equation of a circle $(x - h)^2 + (y - k)^2 = r^2$,where $(h, k)$ is the center.$(x - 2)^2 + (y - (-1))^2 = (\sqrt{50})^2$$(x - 2)^2 + (y + 1)^2 = 50$$x^2 - 4x + 4 + y^2 + 2y + 1 = 50$$x^2 + y^2 - 4x + 2y + 5 = 50$$x^2 + y^2 - 4x + 2y - 45 = 0$.

Find the equation of the circle passing through the point $(3, 6)$ and having its center at $(2, -1)$.

Similar Questions

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

Four distinct points $(2k, 3k), (1, 0), (0, 1)$ and $(0, 0)$ lie on a circle for

The centre and radius of the circle $2x^2 + 2y^2 - x = 0$ are

Find the equation of the circle passing through $(0,0)$ and making intercepts $a$ and $b$ on the coordinate axes.

If each of the lines $5x + 8y = 13$ and $4x - y = 3$ contains a diameter of the circle $x^2 + y^2 - 2(a^2 - 7a + 11)x - 2(a^2 - 6a + 6)y + b^3 + 1 = 0$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module