Find the radical center of the three circles | vedclass

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

Question

(C) Let the equations of the circles be:$S_1: x^2 + y^2 - a^2 = 0$$S_2: x^2 + y^2 - 2cx + c^2 - a^2 = 0$$S_3: x^2 + y^2 - 2by + b^2 - a^2 = 0$The radi

Vedclass · Accepted Answer

$(c/2, b/2)$

Vedclass · Answer

(C) Let the equations of the circles be:$S_1: x^2 + y^2 - a^2 = 0$$S_2: x^2 + y^2 - 2cx + c^2 - a^2 = 0$$S_3: x^2 + y^2 - 2by + b^2 - a^2 = 0$The radical axis of $S_1$ and $S_2$ is given by $S_1 - S_2 = 0$:$(x^2 + y^2 - a^2) - (x^2 + y^2 - 2cx + c^2 - a^2) = 0$$2cx - c^2 = 0 \implies x = c/2$The radical axis of $S_1$ and $S_3$ is given by $S_1 - S_3 = 0$:$(x^2 + y^2 - a^2) - (x^2 + y^2 - 2by + b^2 - a^2) = 0$$2by - b^2 = 0 \implies y = b/2$The radical center is the intersection of these radical axes,which is $(c/2, b/2)$.

Find the radical center of the three circles $x^2 + y^2 = a^2$,$(x - c)^2 + y^2 = a^2$,and $x^2 + (y - b)^2 = a^2$.

Similar Questions

The equation of the circle passing through the points of intersection of the circles $x^2+y^2+4x+6y-12=0$ and $x^2+y^2-6x-4y-12=0$ and cutting the circle $x^2+y^2-4x+4y+8=0$ orthogonally is

If the angle between the circles $x^2+y^2-4x-6y+k=0$ and $x^2+y^2+8x-4y+11=0$ is $\frac{\pi}{3}$,then a value of $k$ is

The equation of the circle passing through the point $(1, 2)$ and through the points of intersection of $x^2 + y^2 - 4x - 6y - 21 = 0$ and $3x + 4y + 5 = 0$ is given by

If the circle $x^2+y^2+4x-6y+c=0$ bisects the circumference of the circle $x^2+y^2-6x+4y-12=0$,then $c$ is equal to

Any circle passes through the point of intersection of the lines $x + \sqrt{3}y = 1$ and $\sqrt{3}x - y = 2$. If it intersects these lines at points $P$ and $Q$,then the angle subtended by the arc $PQ$ at its centre is ............ $^o$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module