The radical centre of the circles x^2+y^2-4x- | vedclass

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

Question

(D) The equations of the given circles are:$S_1: x^2+y^2-4x-6y+5=0$$S_2: x^2+y^2-2x-4y-1=0$$S_3: x^2+y^2-6x-2y=0$The radical axis of $S_1$ and $S_2$ i

Vedclass · Accepted Answer

$\left(\frac{7}{6}, \frac{11}{6}\right)$

Vedclass · Answer

(D) The equations of the given circles are:$S_1: x^2+y^2-4x-6y+5=0$$S_2: x^2+y^2-2x-4y-1=0$$S_3: x^2+y^2-6x-2y=0$The radical axis of $S_1$ and $S_2$ is given by $S_1 - S_2 = 0$:$(x^2+y^2-4x-6y+5) - (x^2+y^2-2x-4y-1) = 0$$-2x - 2y + 6 = 0 \Rightarrow x+y=3$ ... $(i)$The radical axis of $S_2$ and $S_3$ is given by $S_2 - S_3 = 0$:$(x^2+y^2-2x-4y-1) - (x^2+y^2-6x-2y) = 0$$4x - 2y - 1 = 0$ ... (ii)To find the radical centre,solve equations $(i)$ and (ii):From $(i)$,$y = 3 - x$. Substitute into (ii):$4x - 2(3 - x) - 1 = 0$$4x - 6 + 2x - 1 = 0$$6x = 7 \Rightarrow x = \frac{7}{6}$Substituting $x = \frac{7}{6}$ into $(i)$:$y = 3 - \frac{7}{6} = \frac{18-7}{6} = \frac{11}{6}$Thus,the radical centre is $\left(\frac{7}{6}, \frac{11}{6}\right)$.

The radical centre of the circles $x^2+y^2-4x-6y+5=0$,$x^2+y^2-2x-4y-1=0$ and $x^2+y^2-6x-2y=0$ is equal to

Similar Questions

Find the equation of a circle which passes through the point $(1,2)$ and the points of intersection of the circles $x^2+y^2-8x-6y+21=0$ and $x^2+y^2-2x-15=0$.

If the two circles $(x-1)^2+(y-3)^2=r^2$ and $x^2+y^2-8x+2y+8=0$ intersect in two different points,then what can we conclude about $r$?

If two circles touch at $(0,0)$ externally,then their radical axis is

If $2x+y=0$ is the equation of a chord of the circle $x^2+y^2-2x-6y+3=0$,then the circle with this chord as diameter passes through the point

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module