Question

The equation of the circle with center $(1,2)$ and tangent $x+y-5=0$ is

• A. $x^2+y^2+2x-4y+6=0$
• B. $x^2+y^2-2x-4y+3=0$
• C. $x^2+y^2-2x-4y-8=0$
• D. $x^2+y^2-2x-4y+8=0$

Answer 1

The correct option is B $x^2+y^2-2x-4y+3=0$

Coordinates of the centre of the circle $(h,k)=(1,2)$ and equation of the tangent is $x+y-5=0.$

We know that the standard equation of the straight line is $ax+by+c=0.$

Comparing the given equation with the standard equation, we get $a=1,b=1$ and $c=-5.$

We also know that the length of the perpendicular drawn from the centre $(h,k)$ to the given line is equal to the radius of the circle, i.e.

$\left(p\right)=\frac{|ah+bk+c|}{\sqrt{a^2+b^2}}=\frac{|(1\times 1)+(1\times 2)-5|}{\sqrt{{\left(1\right)}^2+{\left(1\right)}^2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$

We also know that the equation of the circle with $(h,k)$ as coordinates of its centre and $r$ as radius is

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

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

$\Rightarrow x^2+1-2x+y^2+4-4y=2$

$\Rightarrow x^2+y^2-2x-4y+3=0$