Question

Radius of the circle ${x^2} + {y^2} + 2x\cos \theta + 2y\sin \theta - 8 = 0$, is
A. $1$
B. $3$
C. $2\sqrt 3 $
D. $\sqrt {10} $

Answer 1

Hint: In this question, we are given the equation of circle ${x^2} + {y^2} + 2x\cos \theta + 2y\sin \theta - 8 = 0$ and we have to calculate the radius of the circle. First step is to compare the equation with the general equation to know the values of $g$, $f$ and $c$. Now, put the required values in the radius formula and formula. Also, one trigonometric identity will be used while solving i.e., ${\sin^2}\theta + {\cos^2}\theta = 1$.

Formula Used:
General equation of circle – ${x^2} + {y^2} + 2gx + 2fy + c = 0$
Radius of circle is $r = \sqrt {{g^2} + {f^2} - c} $
Trigonometric identity, ${\sin^2}\theta + {\cos^2}\theta = 1$

Complete step by step solution:
Given that,
The equation of circle is ${x^2} + {y^2} + 2x\cos \theta + 2y\sin \theta - 8 = 0$
Comparing the above equation with the general equation of circle i.e., ${x^2} + {y^2} + 2gx + 2fy + c = 0$
It implies that, $g = \cos \theta $, $f = \sin \theta $ and $c = - 8$
Now, to calculate the radius of the circle
Using the formula $r = \sqrt {{g^2} + {f^2} - c} $
$r = \sqrt {{{\cos}^2}\theta + {{\sin}^2}\theta - \left( { - 8} \right)} $
Using trigonometric identity, ${\sin^2}\theta + {\cos^2}\theta = 1$
$r = \sqrt {1 + 8} $
$r = \sqrt 9 $
$r = 3$

Option ‘B’ is correct

Note: A circle is a closed curve drawn from a fixed point known as the center, with all points on the curve being the same distance from the center point. ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$ is the equation for a circle with $\left( {h,k} \right)$ center and r radius. This is the equation's standard form. The definition of a circle on the coordinate plane can be expressed using the equation of a circle.