Question

A diameter of the circle ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}$=2x+6y−6 is a chord of another circle with centre (2,1). The radius of this circle C is (in units).

Answer 1

Hint: To solve this question, we use the basic theory of circles. As given circle having equation: ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}$=2x+6y−6. First, we convert the given equation into the standard form of the circle equation. And then calculate its centre and radius so that we calculate radius of the new circle by using this as mentioned in the diagram.

Complete step-by-step answer:
As we know,
Chords length = 2$\sqrt {{{\text{r}}^{\text{2}}}{\text{ - }}{{\text{d}}^{\text{2}}}} $equation of circle is ${\left( {{\text{x - h}}} \right)^{\text{2}}}{\text{ + }}{\left( {{\text{y - k}}} \right)^{\text{2}}}{\text{ = }}{{\text{r}}^{\text{2}}}$

Given, equation of circle:
${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}$-2x-6y+6=0
$\left( {{{\text{x}}^{\text{2}}}{\text{ - 2x + 1}}} \right){\text{ + }}\left( {{{\text{y}}^{\text{2}}}{\text{ - 6y + 9}}} \right){\text{ - 1 - 10 + 4 = 0}}$
${\left( {{\text{x - 1}}} \right)^{\text{2}}}{\text{ + }}{\left( {{\text{y - 3}}} \right)^{\text{2}}}{\text{ = 4}}$
So, c= (1, 3) and r=2
As given, diameter of the circle ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}$=2x+6y−6 is a chord of another circle with centre (2,1).
So, in this case point ${\text{O'}}$ is (1, 3) and length O’A=2
Now, we calculate length OO’
OO’=$\sqrt {{{{\text{(2 - 1)}}}^{\text{2}}}{\text{ + (1 - 3}}{{\text{)}}^{\text{2}}}} $= $\sqrt 5 $
Using Pythagoras theorem,
OA=$\sqrt {{{{\text{(OO')}}}^{\text{2}}}{\text{ + (O'A}}{{\text{)}}^{\text{2}}}} $
      = $\sqrt {{\text{(}}\sqrt {{\text{5}}{{\text{)}}^{\text{2}}}} {\text{ + (2}}{{\text{)}}^{\text{2}}}} $
      =$\sqrt 9 $
 OA(r)=3
Therefore,
The radius of this circle C is 3 (in units).


Note- if we want to construct a circle equation with centre (h,k)and the radius a then the equation of circle is as, ${\left( {{\text{x - h}}} \right)^{\text{2}}}{\text{ + }}{\left( {{\text{y - k}}} \right)^{\text{2}}}{\text{ = }}{{\text{r}}^{\text{2}}}$.which is called the standard form for the equation of a circle. Where r is the radius of the given circle.