The centre and radius of the circle 2x^2 + 2y | vedclass

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

Question

(A) The given equation of the circle is $2x^2 + 2y^2 - x = 0$.Dividing the entire equation by $2$,we get $x^2 + y^2 - \frac{1}{2}x = 0$.Comparing this

Vedclass · Accepted Answer

$\left( \frac{1}{4}, 0 \right)$ and $\frac{1}{4}$

Vedclass · Answer

(A) The given equation of the circle is $2x^2 + 2y^2 - x = 0$.Dividing the entire equation by $2$,we get $x^2 + y^2 - \frac{1}{2}x = 0$.Comparing this with the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $2g = -\frac{1}{2} \implies g = -\frac{1}{4}$,$2f = 0 \implies f = 0$,and $c = 0$.The centre of the circle is $(-g, -f) = \left( \frac{1}{4}, 0 \right)$.The radius $R$ is given by $\sqrt{g^2 + f^2 - c} = \sqrt{\left( -\frac{1}{4} \right)^2 + 0^2 - 0} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.Thus,the centre is $\left( \frac{1}{4}, 0 \right)$ and the radius is $\frac{1}{4}$.

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

Similar Questions

If the sides of a rectangle are given by the equations $x=-2, x=6, y=-2, y=5$,then the equation of the circle,drawn on the diagonal of this rectangle as its diameter,is

If the equation of the circle having its centre in the second quadrant touches the coordinate axes and also the line $\frac{x}{5}+\frac{y}{12}=1$ is $x^2+y^2+2 \lambda x-2 \lambda y+\lambda^2=0$,then $\lambda=$

If the power of the point $(1, 6)$ with respect to the circle $x^2 + y^2 + 4x - 6y - a = 0$ is $-16$,then $a =$

Find the equation of the circumcircle of the triangle formed by the lines $x = 0$,$y = 0$,and $\frac{x}{a} - \frac{y}{b} = 1$.

The equation of a circle which touches the $x$-axis and whose centre is $(1, 2)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module