Consider a circle (x-α)^2+(y-β)^2=50 , where α, β>0 . If circle touches line y+x=0 at point P

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10-1.Circle and System of Circles

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Consider a circle $(x-\alpha)^2+(y-\beta)^2=50$, where $\alpha, \beta>0$. If the circle touches the line $y+x=0$ at the point $P$, whose distance from the origin is $4 \sqrt{2}$ , then $(\alpha+\beta)^2$ is equal to................

A

$103$

B

$102$

C

$55$

D

$100$

(JEE MAIN-2024)

Solution

$ S:(x-\alpha)^2+(y-\beta)^2=50 $
$ C P=r $
$ \left|\frac{\alpha+\beta}{\sqrt{2}}\right|=5 \sqrt{2} $
$ \Rightarrow(\alpha+\beta)^2=100$

Standard 11

Mathematics

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