The equation of the circle with centre at (1, | vedclass

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

Question

(A) The given circle is $x^2 + y^2 + 2y - 3 = 0$. Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 0$ and $f = 1$. The

Vedclass · Accepted Answer

$x^2 + y^2 - 2x + 4y + 3 = 0$

Vedclass · Answer

(A) The given circle is $x^2 + y^2 + 2y - 3 = 0$. Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 0$ and $f = 1$. The centre of this circle is $(-g, -f) = (0, -1)$.The required circle has its centre at $(1, -2)$ and passes through $(0, -1)$.The radius $r$ is the distance between $(1, -2)$ and $(0, -1)$,which is $r = \sqrt{(1 - 0)^2 + (-2 - (-1))^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$,where $(h, k) = (1, -2)$.$(x - 1)^2 + (y + 2)^2 = (\sqrt{2})^2$$x^2 - 2x + 1 + y^2 + 4y + 4 = 2$$x^2 + y^2 - 2x + 4y + 3 = 0$.

The equation of the circle with centre at $(1, -2)$ and passing through the centre of the given circle $x^2 + y^2 + 2y - 3 = 0$ is

Similar Questions

The equation of the circle with the origin as the centre passing through the vertices of an equilateral triangle whose median is of length $3a$ is

If the radius of the circle $x^2 + y^2 - 18x + 12y + k = 0$ is $11$,then $k = $

The equation of the circle with centre $(2,3)$ and touching the line $3x-4y+1=0$ is

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

The polar equation of the circle with centre $\left(2, \frac{\pi}{2}\right)$ and radius $3$ units is :

Vedclass Test Series

Exam Paper Generator

Online Exam Module