Question

The equation of the circle with center $(2,1)$ 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

The correct option is C

$x^2+y^2-4x-2y+4=0$

Explanation for the correct answer :

Given the centre of the circle is $(2,1)$ and the circle touches the line $3x+4y=5$.

The distance between the centre and this line will be the radius.

Calculating the distance between $(2,1)$ and $3x+4y=5$:

The distance formula is

$r=\frac{\left|ax+by+c\right|}{\sqrt{a^2+b^2}}$

Substituting the values we get,

$$\begin{gathered} r=\frac{\left|3x+4y-5\right|}{\sqrt{3^2+4^2}}\left[a=3,b=4,c=5\text{from the equation}\right] \\ \Rightarrow r=\frac{\left|3\left(2\right)+4\left(1\right)-5\right|}{\sqrt{25}}\left[\text{given}x=2,y=1\right] \\ \Rightarrow r=\frac{5}{5} \\ \Rightarrow r=1 \end{gathered}$$

Hence, the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$

$$\begin{gathered} {(x-2)}^2+{(y-1)}^2=r\left[\because h=2,k=1\right] \\ \Rightarrow x^2+y^2-4x-2y+5=1 \\ \Rightarrow x^2+y^2-4x-2y+4=0 \end{gathered}$$

Therefore, the correct answer is Option (C).