Question

If a circle $S(x,y)=0$ touches at the point $(2,3)$ of the line $x+y=5$ and $S(1,2)=–2,$ 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 C $\frac{1}{2}$ units

Desired equation of the circle is ${(x-2)}^2+{(y-3)}^2+\lambda (x+y-5)=0$

At $(1,2)\Rightarrow {(1-2)}^2+{(2-3)}^2+\lambda (1+2-5)=0$

$\Rightarrow 1+1+\lambda (1+2-5)=0$

$\Rightarrow 2-2\lambda =0$

$\Rightarrow \lambda =1$

${(x-2)}^2+{(y-3)}^2+\lambda (x+y-5)=0$ becomes

${(x-2)}^2+{(y-3)}^2+(x+y-5)=0$

$x^2+4-4x+y^2-6y+9+x+y-5=0$

$x^2+y^2-3x-5y+8=0$

Hence, radius$=\sqrt{{\left(\frac{3}{2}\right)}^2+{\left(\frac{5}{2}\right)}^2-8}=\sqrt{\frac{9+25-32}{4}}=\frac{1}{2}$