Question

If one of the diameters of the circle $x^2+y^2–2x–6y+6=0$ is a chord of another circle ‘$C$’ whose center is at $\left(2,1\right)$, then its radius is

Answer 1

Step 1: Diagram for the better understanding

From the diagram, we have a circle ‘$C$’ with coordinates of the center $\left(2,1\right)$. Denote it by $A$.

Now the circle with one of its diameters as the chord of the circle ‘$C$’. Let's call that diameter $BD$ and centered at E.

Step 2: Determination of other required coordinates

For the radius $AB$ of the circle ‘$C$’, we will need the coordinates of the point $E$ and radius $BE$

To determine the coordinates of the center of the smaller circle we have its equation.

Now, we know that the general form of the equation of a circle is $x^2+y^2+2hx+2ky+c=0$, where $\left(h,k\right)$ are the coordinates of the center.

Then we have

$2h=2\text{and}2k=6$

Then the coordinates of the point $E$ are $\left(1,3\right)$

The length of the diameter $BD$ is,

$\begin{array}{rcl}BD& =& 6-2\\ & =& 4\text{units}\end{array}$

The radius $BE$ will be,

$\begin{array}{rcl}BE& =& \frac{BD}{2}\\ & =& 2\text{units}\end{array}$

Step 3: Calculation of the radius of circle C.

First, by the distance formula find $AE$,

$\begin{array}{rcl}AE& =& \sqrt{{\left(1\right)}^2+{\left(-2\right)}^2}\\ & =& \sqrt{5}\text{units}\end{array}$

Now, apply the Pythagoras Theorem in a triangle $ABE$,

$\begin{array}{rcl}\therefore A{B}^2& =& B{E}^2+E{A}^2\\ & \Rightarrow & AB=\sqrt{{\left(2\right)}^2+{\left(\sqrt{5}\right)}^2}\\ & =& \sqrt{4+5}\\ & =& 3\text{units}\end{array}$

Hence, the radius of the circle ‘C’ is $\mathbf{3}\mathbf{}\mathbf{\text{units}}$.