The equation of a circle passing through the | vedclass

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

Question

(C) The general equation of a circle passing through the origin is $x^2 + y^2 + 2gx + 2fy = 0$.For a system of co-axial circles,the radical axis of an

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The general equation of a circle passing through the origin is $x^2 + y^2 + 2gx + 2fy = 0$.For a system of co-axial circles,the radical axis of any two circles must be the same.The given circles are $S_1: x^2 + y^2 - a^2 = 0$ and $S_2: x^2 + y^2 + 2ax - 2a^2 = 0$.The radical axis of $S_1$ and $S_2$ is $S_1 - S_2 = 0$,which gives $2ax - a^2 = 0$,or $x = \frac{a}{2}$.Let the required circle be $S_3: x^2 + y^2 + 2gx + 2fy = 0$.The radical axis of $S_1$ and $S_3$ is $x^2 + y^2 - a^2 - (x^2 + y^2 + 2gx + 2fy) = 0$,which simplifies to $2gx + 2fy + a^2 = 0$.Since the circles are co-axial,this radical axis must be $x = \frac{a}{2}$.Comparing $2gx + 2fy + a^2 = 0$ with $x = \frac{a}{2}$ (or $x - \frac{a}{2} = 0$),we get $f = 0$ and $\frac{2g}{1} = \frac{a^2}{-a/2} = -2a$.Thus,$g = -a$.Substituting $g = -a$ and $f = 0$ into the equation of $S_3$,we get $x^2 + y^2 - 2ax = 0$.

The equation of a circle passing through the origin and co-axial to the circles $x^2 + y^2 = a^2$ and $x^2 + y^2 + 2ax = 2a^2$ is

Similar Questions

If the angle between the circles $x^2+y^2-2x+2y+1=0$ and $x^2+y^2+2x-2y+k=0$ is $\frac{\pi}{3}$,then

The centre of the circle which intersects the circle $x^2+y^2-2x-2y-2=0$ orthogonally,passes through the point $(2,0)$,and touches the $X$-axis is:

If the radical axis of the circles $x^2+y^2+2 \alpha x+2 \beta y+c=0$ and $x^2+y^2+\frac{3}{2} x+4 y+c=0$ touches the circle $x^2+y^2+2 x+2 y+1=0$,then $4 \alpha \beta-8 \alpha-3 \beta+10=$

The circles $x^2 + y^2 - 2x - 4y = 0$ and $x^2 + y^2 - 8y - 4 = 0$:

The circle $S=0$ cuts the circle $x^2+y^2-4x+2y-7=0$ orthogonally. If $(2,3)$ is the centre of the circle $S=0$,then its radius is

Vedclass Test Series

Exam Paper Generator

Online Exam Module