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 a circle with center at $(-2, 3)$ and circumference of $4 \pi$ units is

The equation of a circle passing through the origin and making an $x$-intercept of $3$ and a $y$-intercept of $-5$ is

Let the equation $ax^2+2hxy+by^2+2gx+2fy+c=0$ represent a point circle other than the origin. Then which one of the following conditions must hold?

If the line $3x - 2y + 6 = 0$ meets the $X$-axis and $Y$-axis at points $A$ and $B$ respectively,then the equation of the circle with radius $AB$ and center at $A$ is

The number of circles touching the lines $x = 0$,$y = a$ and $y = b$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module