Exam

JEE Main

Year

2019

Date

10 April

Shift

Morning Shift

Subject

Mathematics

Chapter

Circle

JEE Main 2019 · 10 April · Morning Shift · Mathematics · Circle

The line x = y touches a circle at the point ( 1, 1 ) . If the...

The line $x = y$ touches a circle at the point $( 1, 1 )$. If the circle also passes through the point $( 1, - 3 )$, then its radius is

Step-by-step solution

Equation of family of circles touches line $x – y = 0$ at point $( 1,1 )$ is
${( x - 1 )}^{2} + {( y - 1 )}^{2} + \lambda\left(x – y\right) = 0, \lambda \in R$
If a member of this family, also passes through point $( 1, – 3 )$ then
${\left(1 - 1\right)}^{2} + {\left(- 3 - 1\right)}^{2} + \lambda\left(1 - \left(- 3\right)\right) = 0 ⇒ \lambda = - 4$
$⇒$ Required circle is
${\left(x - 1\right)}^{2} + {\left(y - 1\right)}^{2} - 4\left(x - y\right) = 0$
$⇒$ ${x}^{2} + {y}^{2} - 6x + 2y + 2 = 0$
Where radius is $r = \sqrt{{3}^{2} + {1}^{2} - 2} = 2\sqrt{2}$