Question

Equation of the circle, centred at $(1,–5)$ and touching the line $3x+4y=8,$ is

• A. $x^2+y^2-2x+10y+1=0$
• B. $x^2+y^2-4x+5y+25=0$
• C. $x^2+y^2+2x-10y+26=0$
• D. $x^2+y^2-10x+2y+1=0$

Answer 1

The correct option is A $x^2+y^2-2x+10y+1=0$

Let $r$ be the radius of required circle.
Then $r$ is perpendicular distance of point $(1,-5)$ from $3x+4y-8=0$
$\Rightarrow r=\frac{|3\left(1\right)+4(-5)-8|}{\sqrt{3^2+4^2}}\Rightarrow r=5$
Hence required equation of circle is,
$(x-1{)}^2+(y+5{)}^2=5^2\Rightarrow x^2+y^2-2x+10y+1=0$