The equation of the radical axis of the circl | vedclass

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

Question

(B) The equation of the radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$,provided the coefficients of $x^2$ and $y^2$ a

Vedclass · Accepted Answer

$7x - 3y + 18 = 0$

Vedclass · Answer

(B) The equation of the radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$,provided the coefficients of $x^2$ and $y^2$ are normalized to $1$.Given circles:$S_1: x^2 + y^2 + x - y + 2 = 0$$S_2: 3x^2 + 3y^2 - 4x - 12 = 0$Divide $S_2$ by $3$ to normalize:$S_2': x^2 + y^2 - \frac{4}{3}x - 4 = 0$Now,find $S_1 - S_2' = 0$:$(x^2 + y^2 + x - y + 2) - (x^2 + y^2 - \frac{4}{3}x - 4) = 0$$x + \frac{4}{3}x - y + 2 + 4 = 0$$\frac{7}{3}x - y + 6 = 0$Multiply by $3$ to simplify:$7x - 3y + 18 = 0$

The equation of the radical axis of the circles $x^2 + y^2 + x - y + 2 = 0$ and $3x^2 + 3y^2 - 4x - 12 = 0$ is:

Similar Questions

Find the equation of the circle passing through the intersection of the circles $x^{2} + y^{2} - 8x - 2y + 7 = 0$ and $x^{2} + y^{2} - 4x + 10y + 8 = 0$,and having its center on the $y-$axis.

Find the equation of the circle which cuts orthogonally each of the three circles $x^2+y^2-2x+3y-7=0$,$x^2+y^2+5x-5y+9=0$,and $x^2+y^2+7x-9y+29=0$.

If $y = 2x$ is a chord of the circle $x^{2} + y^{2} = 10x$,then find the equation of the circle having this chord as its diameter.

From which point are the lengths of the tangents drawn to the circles $x^{2} + y^{2} - 8x + 40 = 0$,$5x^{2} + 5y^{2} - 25x + 80 = 0$,and $x^{2} + y^{2} - 8x + 16y + 160 = 0$ equal?

The radical centre of the circles $x^2+y^2-4x-6y+5=0$,$x^2+y^2-2x-4y-1=0$ and $x^2+y^2-6x-2y=0$ lies on the line

Vedclass Test Series

Exam Paper Generator

Online Exam Module