The length of the common chord of the circles | vedclass

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

Question

(A) The equations of the circles are:$S_1: (x - a)^2 + (y - b)^2 = c^2$$S_2: (x - b)^2 + (y - a)^2 = c^2$Subtracting $S_2$ from $S_1$ gives the equati

Vedclass · Accepted Answer

$\sqrt{4c^2 - 2(a - b)^2}$

Vedclass · Answer

(A) The equations of the circles are:$S_1: (x - a)^2 + (y - b)^2 = c^2$$S_2: (x - b)^2 + (y - a)^2 = c^2$Subtracting $S_2$ from $S_1$ gives the equation of the common chord:$(x - a)^2 - (x - b)^2 + (y - b)^2 - (y - a)^2 = 0$$(x^2 - 2ax + a^2) - (x^2 - 2bx + b^2) + (y^2 - 2by + b^2) - (y^2 - 2ay + a^2) = 0$$-2ax + 2bx - 2by + 2ay = 0$$-2a(x - y) + 2b(x - y) = 0$$(2b - 2a)(x - y) = 0$Since $a 
eq b$,we have $x - y = 0$,or $y = x$.Substitute $y = x$ into the first circle equation:$(x - a)^2 + (x - b)^2 = c^2$$x^2 - 2ax + a^2 + x^2 - 2bx + b^2 = c^2$$2x^2 - 2(a + b)x + (a^2 + b^2 - c^2) = 0$The distance $d$ from the center $(a, b)$ to the line $x - y = 0$ is:$d = \frac{|a - b|}{\sqrt{1^2 + (-1)^2}} = \frac{|a - b|}{\sqrt{2}}$The length of the common chord is $2\sqrt{r^2 - d^2}$:$= 2\sqrt{c^2 - \frac{(a - b)^2}{2}}$$= 2\sqrt{\frac{2c^2 - (a - b)^2}{2}}$$= \sqrt{2(2c^2 - (a - b)^2)}$$= \sqrt{4c^2 - 2(a - b)^2}$

The length of the common chord of the circles $(x - a)^2 + (y - b)^2 = c^2$ and $(x - b)^2 + (y - a)^2 = c^2$ is

Similar Questions

Find the distance from the center of the circle $x^2 + y^2 = 2x$ to the common chord of the circles $x^2 + y^2 + 5x - 8y + 1 = 0$ and $x^2 + y^2 - 3x + 7y - 25 = 0$.

The equation of the circle whose diameter is the common chord of the circles $x^2+y^2+2x+2y+1=0$ and $x^2+y^2+4x+6y+4=0$ is

$A$ 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 circle $S \equiv x^2+y^2-2x-4y+1=0$ cuts the $y$-axis at $A, B$ $(OA > OB)$. If the radical axis of $S=0$ and $S^{\prime} \equiv x^2+y^2-4x-2y+4=0$ cuts the $y$-axis at $C$,then the ratio in which $C$ divides $AB$ is:

For the circles $(x-a)^2+y^2=a^2$ and $x^2+(y-a)^2=a^2$,where $a>0$,which one of the following is not true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module