Question

A circle has radius $3$ units and its center lies on the line $y = x - 1$. If it passes through $\left( {7,3} \right)$, its equation is
A. ${x^2} + {y^2} - 8x - 14y = 0$
B. ${x^2} + {y^2} - 8x - 6y - 6 = 0$
C. ${x^2} + {y^2} - 14x - 12y - 76 = 0$
D. ${x^2} + {y^2} + 14x + 12y - 70 = 0$

Answer 1

Hint: In this question, we are given the radius and the equation of line passing through the center. Also, the point which is passing through the circle. First step is to let the center be $\left( {h,k} \right)$and use the standard form of circle to find its values by taking $k = h - 1$( we converted this because the line is passing through the center. Lastly put the value of center and radius in standard form.

Formula Used:
Equation of Circle (Standard form) –
${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$, here $\left( {h,k} \right)$ is the center and $r$ is the radius of the circle.

Complete step by step solution:
Given that,
Radius of the circle is $r = 3$ units and the circle is passing through the points $\left( {x,y} \right)$ be $\left( {7,3} \right)$
Let, the center of the circle be $\left( {h,k} \right)$
Therefore, the line passing through the center will convert from $y = x - 1$ to $k = h - 1 - - - - - \left( 1 \right)$
Now, the standard equation of the circle is ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2} - - - - - \left( 2 \right)$
Circle is passing through $\left( {7,3} \right)$. Put $x = 7$ and $y = 3$ in equation (2),
${\left( {7 - h} \right)^2} + {\left( {3 - k} \right)^2} = {3^2}$
${\left( {7 - h} \right)^2} + {\left( {3 - \left( {h - 1} \right)} \right)^2} = {3^2}$
$49 + {h^2} - 14h + 16 + {h^2} - 8h = 9$
$2{h^2} - 22h + 56 = 0$
${h^2} - 11h + 28 = 0$
$\left( {h - 7} \right)\left( {h - 4} \right) = 0$
$h = 7,h = 4$
For, $h = 7 \Rightarrow k = 6$
And $h = 4 \Rightarrow k = 3$
From equation (2)
At center $\left( {7,6} \right)$, Equation of the circle is ${x^2} + {y^2} - 14x - 12y - 76 = 0$
And at $\left( {4,3} \right)$. Equation of the circle is ${x^2} + {y^2} - 8x - 6y + 16 = 0$

Option ‘C’ 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. While solving such questions always keep in mind to find the equation we need the value of the radius and the point of center. Also, one should always remember each and every equation of circle (General and standard both). Also, the formulas to find the radius and the distance between any point and line.