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 radius of the circle is:

• A. $3$
• B. $2$
• C. $\frac{3}{2}$
• D. $1$

Answer 1

The correct option is A $3$

Equation of the circle is given by,

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

$\therefore (x^2-2x)+(y^2-6y)+6=0$

$\therefore (x^2-2x+1)-1+(y^2-6y+9)-9+6=0$

$\therefore {(x-1)}^2+{(y-3)}^2=9+1-6$

$\therefore {(x-1)}^2+{(y-3)}^2=4$

$\therefore {(x-1)}^2+{(y-3)}^2={\left(2\right)}^2$

Comparing this equation with standard form of the equation i.e. $\therefore {(x-h)}^2+{(y-k)}^2={\left(r\right)}^2$, we get,

$h=1$, $k=3$ and $r=2$

Thus, coordinates of center of circle are, $C(1,3)$

Let AB is diameter of this circle. C will be mid-point of this diameter.

As given in the problem, AB will be chord of another circle having center $D(2,1)$

By property of chord of circle, as C is mid-point of chord AB, $AB\perp CD$

As shown in figure, AC = 2

By distance formula, $CD=\sqrt{{(2-1)}^2+{(1-3)}^2}$

$\therefore CD=\sqrt{{\left(1\right)}^2+{(-2)}^2}$

$\therefore CD=\sqrt{1+4}=\sqrt{5}$

In right angled triangle $\Delta ACD$, By Pythagoras theorem,

${AD}^2={AC}^2+{CD}^2$

$\therefore {AD}^2={\left(2\right)}^2+{\left(\sqrt{5}\right)}^2$

$\therefore {AD}^2=4+5$

$\therefore {AD}^2=9$

Taking square roots of both sides, we get,

$AD=3$

But, as shown in figure, AD = radius of another circle having radius $(2,1)$

Thus, Radius of circle with AB as chord is 3.

Thus, answer is option (A)