Question

The equation to the circle with center $\left( {2,1} \right)$ and touching the line $3x + 4y = 5$ is
A. ${x^2} + {y^2} - 4x - 2y + 5 = 0$
B. ${x^2} + {y^2} - 4x - 2y - 5 = 0$
C. ${x^2} + {y^2} - 4x - 2y + 4 = 0$
D. ${x^2} + {y^2} - 4x - 2y - 4 = 0$

Answer 1

Hint: In this question, we are given the centre of the circle and also the line $3x + 4y = 5$ which is touching the circle. We have to find the equation of the circle. First step is to find the radius of the circle as we have to find the equation ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$ where $\left( {h,k} \right)$is the centre and $r$ is radius. To calculate the radius, use the distance formula $D = \dfrac{{\left| {a{x_0} + b{y_0} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$, here it is the distance between the point $\left( {{x_0},{y_0}} \right)$ and the line $ax + by + c = 0$.

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
The distance of the line from the point $\left( {{x_0},{y_0}} \right)$ to the line $ax + by + c = 0$ –
 $D = \dfrac{{\left| {a{x_0} + b{y_0} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$

Complete step by step solution:
Given that,
Coordinate of the center of circle are $\left( {2,1} \right)$and the circle is touching the line $3x + 4y = 5$
As we know that,

The distance from the line $ax + by + c = 0$ to the point $\left( {{x_0},{y_0}} \right)$ is $D = \dfrac{{\left| {a{x_0} + b{y_0} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$
Here, the distance between the line and the center is the radius of the circle.
Therefore, $r = \dfrac{{\left| {3\left( 2 \right) + 4\left( 1 \right) + \left( { - 5} \right)} \right|}}{{\sqrt {{3^2} + {4^2}} }}$
$r = \dfrac{{\left| {6 + 4 - 5} \right|}}{{\sqrt {9 + 16} }}$
$r = \dfrac{{\left| 5 \right|}}{{\sqrt {25} }}$
$r = 1$
Now, using the standard form of the equation of circle is
${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$
${\left( {x - 2} \right)^2} + {\left( {y - 1} \right)^2} = {\left( 1 \right)^2}$
${x^2} + {2^2} - 2\left( 2 \right)\left( x \right) + {y^2} + {1^2} - 2\left( 1 \right)\left( y \right) = 1$
${x^2} + {y^2} - 4x - 2y + 4 = 0$

Option ‘C’ is correct

Note: To solve such a question, first try to make the figure and understand the question properly. One should always remember each and every equation of the circle (General and standard both). Also, the formulas to find the radius and the distance between any point and line. Don’t get confused by the distance formula. There are two distance formulas one which we used in this question, and another is $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $which we use to find the distance of line from one point to the other.