Question

If one of the diameters of the circle $x^2+y^2-2x-6y+6=0$ is a chord to the circle with centre $(2,1)$, then the equation of the circle is

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

Answer 1

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

According to the given question


The given circle is
$x^2+y^2-2x-6y+6=0$
$C=(-g,-f)=(1,3)$
$r=\sqrt{g^2+f^2-c}=\sqrt{1+9-6}=2$

Let $AB$ be one of its diameter which is the chord of other circle with center $C_1(2,1)$, so in $△C_{1}CB$, we have
$(C_{1}B{)}^2=(C{C}_1{)}^2+(CB{)}^2\Rightarrow r^2=[(2-1{)}^2+(1-3{)}^2]+2^2$
$\Rightarrow r=3$

Hence, the equation of the other circle is
$(x-2{)}^2+(y-1{)}^2=3^2\Rightarrow x^2+y^2-4x-2y-4=0$