Question

If one of the diameters of the circle, given by the equation, ${x^2} + {y^2} - 4x + 6y - 12 = 0$ is a chord of circle S, whose centre is at $\left( { - 3,2} \right)$, then the radius of S is:
(A) $5\sqrt 2$
(B) $5\sqrt 3$
(C) $5$
(D) $10$

Answer 1

Hint:Firstly, Compare the given circle ${x^2} + {y^2} - 8x - 8y - 4 = 0$ with the general equation a circle, i.e., ${x^2} + {y^2} + 2gx + 2fy + c = 0$ and find out the centre of circle $\left( { - g, - f} \right)$ and radius of circle $\sqrt {{g^2} + {f^2} - c}$. Then draw the diagram according to given information and find out the required value.

Complete step-by-step answer:

Given circle is ${x^2} + {y^2} - 4x + 6y - 12 = 0$

Compare this given equation with the general equation a circle i.e., ${x^2} + {y^2} + 2gx + 2fy + c = 0$, we get-

$2g = - 4 \Rightarrow g = - 2$

$2f = 6 \Rightarrow f = 3$

$c = - 12$

Centre of circle $= \left( { - g, - f} \right) = \left( {2, - 3} \right)$

Radius of circle $= \sqrt {{g^2} + {f^2} - c} = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 3} \right)}^2} - \left( { - 12} \right)} = \sqrt {4 + 9 + 12} = \sqrt {25} = 5$

We have given that one of the diameters of the given circle is a chord of circle S, whose center is at $\left( { - 3,2} \right)$.

Now, according to given information, we have the following figure:

Clearly, $AO \bot BC$, as $O$ is mid-point of the chord.

Now, in $\vartriangle AOB$, $OA$ is find by applying the distance formula between two points $\left( {{x_1},{y_1}} \right)$and $\left( {{x_2},{y_2}} \right)$ i.e., $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$.

$\therefore$ $OA = \sqrt {{{\left( { - 3 - 2} \right)}^2} + {{\left( {2 + 3} \right)}^2}} = \sqrt {25 + 25} = \sqrt {50} = 5\sqrt 2$

And $OB = 5$ (Radius of the given circle)

Now applying Pythagoras theorem in right angle triangle$\vartriangle AOB$, i.e.In a right angled triangle, the square of the hypotenuse is equal to the sum of squares of rest two sides.

Therefore, ${\left( {AB} \right)^2} = {\left( {OA} \right)^2} + {\left( {OB} \right)^2}$

$\Rightarrow {\left( {AB} \right)^2} = {\left( {5\sqrt 2 } \right)^2} + {\left( 5 \right)^2}$

$\Rightarrow {\left( {AB} \right)^2} = 50 + 25$

$\Rightarrow {\left( {AB} \right)^2} = 75$

$\Rightarrow AB = \sqrt {75}$

$\Rightarrow AB = 5\sqrt 3$

Hence the radius of circle S is $5\sqrt 3$.

So, the correct answer is “Option B”.

Note:The Pythagoras theorem states that in a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of other two sides. Mathematically,${\left( {Hypotenuse} \right)^2} = {\left( {Base} \right)^2} + {\left( {Perpendicular} \right)^2}$.Students should remember general equation of circle i.e ${x^2} + {y^2} + 2gx + 2fy + c = 0$, centre $\left( { - g, - f} \right)$ and radius of circle $\sqrt {{g^2} + {f^2} - c}$ for solving these types of questions.