Question

Common chord of the circles $x^2+y^2-6x-4y+9=0$ and $x^2+y^2-8x-6y+23=0$ is a diameter of the second circle then its length is

• A. $4\sqrt{2}$
• B. $\sqrt{2}$
• C. $3\sqrt{2}$
• D. $2\sqrt{2}$

Answer 1

The correct option is D $2\sqrt{2}$

The equation of the common chord is

$S_2-S_1=0$

$(x^2+y^2-6x-4y+9)-(x^2+y^2-8x-6y+23)=0$

$2x+2y-14=0$

$x+y=7$...(i)

Now this is the diameter of $x^2+y^2-8x-6y+23=0$

Hence it should cut the circle at two distinct points.

Therefore

$x^2+(7-x{)}^2-8x-6(7-x)+23=0$
$x^2+x^2-14x+49-8x-42+6x+23=0$

$2x^2+x(-14-8+6)+49-42+23=0$

$2x^2-16x+30=0$

$x^2-8x+15=0$

$x=5,3$

$y=2,4$ repectively.

$D=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$

$=\sqrt{2^2+2^2}$

$=2\sqrt{2}$.