Question

The equation to the circle with centre $(2,1)$ 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 $x^2+y^2-4x-2y+4=0$

The distance from the line $ax+by+c=0$ from the line to a point $(x_0,y_0)$ is:

$D=\frac{|a{x}_0+b{y}_0+c|}{\sqrt{a^2+b^2}}$

Here, the distance is equal to the radius of the circle and the distance from the line $3x+4y-5=0$ to a point $(2,1)$ is:

$r=\frac{\left|\right(3\times 2)+(4\times 1)+(-5\left)\right|}{\sqrt{3^2+4^2}}=\frac{|6+4-5|}{\sqrt{9+16}}=\frac{\left|5\right|}{\sqrt{25}}=\frac{5}{5}=1$

Therefore, equation of the circle with centre $(2,1)$ and radius $1$ is:

${(x-2)}^2+{(y-1)}^2={\left(1\right)}^2\Rightarrow x^2+4-4x+y^2+1-2y=1\Rightarrow x^2+y^2-4x-2y+5=1\Rightarrow x^2+y^2-4x-2y+4=0$

Hence, the equation of circle is $x^2+y^2-4x-2y+4=0$.