Question

The equation of the diameter of circle $x^2+y^2+2x-4y-11=0$ which bisects the chords intercepted on the line $2x-y+3=0$ is

• A. $x+y-7=0$
• B. $2x-y-5=0$
• C. $x+2y-3=0$
• D. $Noneofthese$

Answer 1

The correct option is C $x+2y-3=0$

$x^2+y^2+2x-4y-11=(x+1{)}^2+(y-2{)}^2-1-4-11=(x+1{)}^2+(y-2{)}^2=16$

replace $x$ by $x^{\prime }-1$,$y$ by $y^{\prime }+2$.

$x^{\prime 2}+y^{\prime 2}=16$

Given line equation is $2x-y+3=0$

replacing x by x'-1, y by y'+2 ,

$2(x^{\prime }-1)-(y^{\prime }+2)+3=0=2x^{\prime }-y^{\prime }-1=0$

$y^{\prime }=2x^{\prime }-1$

Substituing this is in circle equation,

$x^{\prime 2}+(2x^{\prime }-1{)}^2=16$

$5x^{\prime 2}-4x^{\prime }-15=0$

Let solutions to this eq'n be $x_1,x_2$

$x_1+x_2=\left(4/5\right),y_1+y_2=2x_1-1+2x_2-1=\left(8/5\right)-2=(-2/5)$

mid point of $(x_1,y_1),(x_2,y_2)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(2/5,-1/5)$

Given the line segment passes through the midpoint and the center of the circle $(0,0)$.

equation of the line is $-2y^{\prime }=x^{\prime }$

$-2(y-2)=x+1$

$x+2y=3$