Two circles centered at (2,3) and (5,6) inter | vedclass

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

Question

(C) Let the radius of both circles be $r$. The equation of the circle with center at $(2,3)$ is: $S_{1} \equiv (x-2)^{2} + (y-3)^{2} = r^{2} \quad \do

Vedclass · Accepted Answer

$x+y-8=0$

Vedclass · Answer

(C) Let the radius of both circles be $r$. The equation of the circle with center at $(2,3)$ is: $S_{1} \equiv (x-2)^{2} + (y-3)^{2} = r^{2} \quad \dots(i)$ The equation of the circle with center at $(5,6)$ is: $S_{2} \equiv (x-5)^{2} + (y-6)^{2} = r^{2} \quad \dots(ii)$ The equation of the common chord is given by the radical axis $S_{1} - S_{2} = 0$. $(x-2)^{2} + (y-3)^{2} - ((x-5)^{2} + (y-6)^{2}) = 0$ $(x^{2} - 4x + 4 + y^{2} - 6y + 9) - (x^{2} - 10x + 25 + y^{2} - 12y + 36) = 0$ $(x^{2} + y^{2} - 4x - 6y + 13) - (x^{2} + y^{2} - 10x - 12y + 61) = 0$ $6x + 6y - 48 = 0$ Dividing by $6$,we get: $x + y - 8 = 0$

Two circles centered at $(2,3)$ and $(5,6)$ intersect each other. If the radii are equal,the equation of the common chord is

Similar Questions

$A$ line $lx + my + n = 0$ meets the circle $x^2 + y^2 = a^2$ at the points $P$ and $Q$. The tangents drawn at the points $P$ and $Q$ meet at $R$. Then,the coordinates of $R$ are:

The length of the common chord of the circles $x^2 + y^2 + 2x + 4y - 20 = 0$ and $x^2 + y^2 + 6x - 8y + 10 = 0$ is

If $\frac{2}{\sqrt{5}}$ is the length of the common chord of the circles $x^2+y^2+2x+2y+1=0$ and $x^2+y^2+\alpha x+3y+2=0, \alpha \neq 0$,then $\alpha=$

If $A, B$ are the points of contact of the tangents drawn from the point $P(-2, -3)$ to the circle $x^2+y^2-8x-10y+5=0$ and the chord $AB$ subtends an angle $\theta$ at $P$,then $\tan \theta =$

The equation of the circle whose diameter is the common chord of the circles $x^2+y^2+2x+3y+2=0$ and $x^2+y^2+2x-3y-4=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module