Question

The diameter of a circle are along $2x+y-7=0$ and $x+3y-11=0$. Then the equation of this circle which also passes through $(5,7)$ is

• A. $x^2+y^2-4x-6y-16=0$
• B. $x^2+y^2-4x-6y-20=0$
• C. $x^2+y^2-4x-6y-12=0$
• D. $x^2+y^2+4x+6y-12=0$

Answer 1

The correct option is C $x^2+y^2-4x-6y-12=0$

Given two diameters of circle

$2x+y-7=0$

$x+3y-11=0$

Given circle passes through point $(5,7)$

Lets assume circle equation is $(x-h{)}^2+(y-k{)}^2=r^2$

$(h,k)\to$ centre

$r\to$ radius

By solving the diameters we get center as diameters passes through centre

$2x+y-7=02x+6y-22=0\_\_\_\_\_\_\_\_\_\_\_\_\_\_-5y+15=0\Rightarrow y=3$

$2x+3-7=0$

$x-2$

$\therefore$ centre $(2,3)$

point on circle $=(5,7)$

radius distance between centre and point

$\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}=\sqrt{(5-2{)}^2+(7-3{)}^2}$

$=\sqrt{25}=5$

By substituting centre $(2,3)$ radius $5$

in circle equation

$(x-2{)}^2+(y-3{)}^2=5^2$

$x^2+y^2-4x-6y+13=25$

$x^2+y^2-4x-6y-12=0$