Question

The equation of a circle with centre $(4,1)$ and having $3 x+4 y-1=0$ as tangent, is

(a) $x^2+y^2-8 x-2 y-8=0$

(b) $x^2+y^2-8 x-2 y+8=0$

(c) $x^2+y^2-8 x+2 y-8=0$

(d) $x^2+y^2-8 x-2 y+4=0$

Answer 1

Here, centre $C=(4,1)$

$r=$ Perpendicular distance from the centre to the tangent

$=\frac{4 \cdot 3+4 \cdot 1-1}{\sqrt{3^2+4^2}}=\frac{12+4-1}{\sqrt{25}}=3$

$\therefore$ Equation of circle is

$(x-4)^2+(y-1)^2=9 \Rightarrow x^2+y^2-8 x-2 y+8=0$

So, The correct option of this question will be (B).