If (alpha, beta) is the centre of a circle pa | vedclass

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

Question

(C) The centre of the circle is $(\alpha, \beta)$ and it passes through the origin $(0, 0)$.The radius $r$ is the distance between the centre $(\alpha

Vedclass · Accepted Answer

$x^2 + y^2 - 2\alpha x - 2\beta y = 0$

Vedclass · Answer

(C) The centre of the circle is $(\alpha, \beta)$ and it passes through the origin $(0, 0)$.The radius $r$ is the distance between the centre $(\alpha, \beta)$ and the origin $(0, 0)$:$r = \sqrt{(\alpha - 0)^2 + (\beta - 0)^2} = \sqrt{\alpha^2 + \beta^2}$.The standard equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.Substituting $h = \alpha$,$k = \beta$,and $r^2 = \alpha^2 + \beta^2$:$(x - \alpha)^2 + (y - \beta)^2 = \alpha^2 + \beta^2$Expanding the squares:$x^2 - 2\alpha x + \alpha^2 + y^2 - 2\beta y + \beta^2 = \alpha^2 + \beta^2$Simplifying the equation:$x^2 + y^2 - 2\alpha x - 2\beta y = 0$.

If $(\alpha, \beta)$ is the centre of a circle passing through the origin,then its equation is

Similar Questions

The equation of the circle which has its centre at the point $(3, 4)$ and touches the line $5x + 12y - 11 = 0$ is

Find the equation of the circle with centre $(-3, 2)$ and radius $4$.

$A$ 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

The equation of the circle with radius $5$ and touching the coordinate axes in the third quadrant is:

Find the equation of a circle whose radius is $5$ units and passes through two points on the $x$-axis which are at a distance of $4$ units from the origin.

Vedclass Test Series

Exam Paper Generator

Online Exam Module