The equation of a circle that passes through | vedclass

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

Question

(C) The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$.Since the circle passes through the origin $(0,0)$,we have $c=0$.The circle cuts off an

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$.Since the circle passes through the origin $(0,0)$,we have $c=0$.The circle cuts off an intercept of $-2$ on the $X$-axis,so it passes through $(-2,0)$. Substituting this into the equation: $(-2)^2 + 0^2 + 2g(-2) + 2f(0) + 0 = 0$ $\Rightarrow 4 - 4g = 0$ $\Rightarrow g = 1$.The circle cuts off an intercept of $3$ on the $Y$-axis,so it passes through $(0,3)$. Substituting this into the equation: $0^2 + 3^2 + 2g(0) + 2f(3) + 0 = 0$ $\Rightarrow 9 + 6f = 0$ $\Rightarrow f = -\frac{3}{2}$.Substituting $g=1$ and $f=-\frac{3}{2}$ into the general equation: $x^2+y^2+2(1)x+2(-\frac{3}{2})y+0=0$.This simplifies to $x^2+y^2+2x-3y=0$.

The equation of a circle that passes through the origin and cuts off intercepts $-2$ and $3$ on the $X$-axis and $Y$-axis respectively is

Similar Questions

If the circle $x^{2}+y^{2}-2gx+6y-19c=0$,where $g, c \in R$,passes through the point $(6,1)$ and its centre lies on the line $x-2cy=8$,then the length of the intercept made by the circle on the $x$-axis 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 $\lambda=$

Find the equation of a circle with radius $4$ that touches the $x$-axis at a distance of $-3$ from the origin.

The equations of the circles touching both the axes and passing through the point $(1, 2)$ are

The equation of the circle with centre on the $x$-axis,radius $4$ and passing through the origin,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module