Question

The equation of the circle with centre on the x-axis and touching the line $3x+4y-11=0$ at the point $(1,2)$ is

• A. $x^2+y^2-x-4=0$
• B. $x^2+y^2+2x-7=0$
• C. $x^2+y^2+x-6=0$
• D. none of these

Answer 1

The correct option is C $x^2+y^2+x-6=0$

Let $(a,0)$ be the center and $r$ the radius

The equation will be

$(x-a{)}^2+y^2=r^2$

It passes through $(1,2)$

$(1-a{)}^2+4=r^2$

And since $3x+4y–11=0$ is a tangent

Radius = perpendicular distance from center to the line

$⟹r=\frac{|3a+0-11|}{\sqrt{3^2+4^2}}$

$⟹(5r{)}^2=(3a-11{)}^2$

$⟹25\left(\right(1-a{)}^2+4)=(3a-11{)}^2$

$⟹16a^2+16a+4=0$

$⟹4a^2+4a+1=0$

$a=\frac{-1}{2}⟹r=\frac{5}{2}$

Equation is

$(x+\frac{1}{2}{)}^2+y^2=\frac{25}{4}$

$⟹x^2+y^2+x=6$