Question

If a circle $S(x,y)=0$ touches at the point $(2,3)$ of the line $x+y=5$ and $S(1,2)=0$, then radius of such circle

• A. $2$ units
• B. $4$ units
• C. $\frac{1}{2}$ units
• D. $\frac{1}{\sqrt{2}}$ units

Answer 1

The correct option is D $\frac{1}{\sqrt{2}}$ units

Let $(x-a{)}^2+(y-b{)}^2=r^2$ be the equation of circle with center (a,b) and radius r

Given $S(1,2)=0$

$⟹(1-a{)}^2+(2-b{)}^2–r^2=0-eq.1$

It also passes through (2,3)

$⟹(2-a{)}^2+(3-b{)}^2–r^2=0-eq.2$

Eq.1 = eq.2

$⟹(1-a{)}^2+(2-b{)}^2=(2-a{)}^2+(3-b{)}^2$

$⟹(2a-5)=(5–2b)$

$⟹a+b=4$

And given $x+y–5=0$ is a tangent

$⟹$ radius = perpendicular distance from center to tangent

$r=\frac{|a+b-5|}{\sqrt{1+1}}$

$r=\frac{|4-5|}{\sqrt{2}}=\frac{1}{\sqrt{2}}$