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

Option: 1

\mathrm{2 \, units}


Option: 2

\mathrm{4 \, units}


Option: 3

\mathrm{1/2\: units}


Option: 4

\frac{1}{\sqrt{2}}\: \text{units}


Answers (1)

best_answer

Desired equation of the circle is

(x-2)^{2}+(y-3)^{2}+\lambda(x+y-5)=0
1+1+\lambda(1+2-5)=0

\Rightarrow \lambda=1
x^{2}-4 x+4+y^{2}-6 y+9+x+y-5=0
\Rightarrow x^{2}+y^{2}-3 x-5 y+8=0

\left(x^{2}+\frac{3}{2}\right)^{2}+\left(y-\frac{5}{2}\right)^{2}=-8+\frac{25}{4}+\frac{9}{4}=\frac{2}{4}=\frac{1}{2}

Hence (\mathrm{d}) is the correct answer.
 

Posted by

vishal kumar

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