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

Centre: (2, 1)
Radius is distance of point (2, 1) from 3x + 4y = 5

i.e., $\left|\frac{3.2+4.1-5}{\sqrt{9+16}}\right|\Rightarrow \left|\frac{6+4-5}{5}\right|=1$

$\therefore$ circle is $(x-2{)}^2+(y-1{)}^2=1$
$x^2+y^2-4x-2y+4=0.$