Consider a circle x-α^2+y-β^2=50, where α,β0. If the circle touches the line y+x=0 at the point P, whose distance from the origin is 4√2 , then…

Question

Consider a circle x-α^2+y-β^2=50, where α,β>0. If the circle touches the line y+x=0 at the point P, whose distance from the origin is 4√2 , then (α+β)^2 is equal to _______.

MARKS Content Team · Accepted Answer

Correct answer: 100. Given: x-α^2+y-β^2=50 The given equation of circle represent centre as α,β and radius as 5√2 units. Now, x+y=0 is tangent to the given circle at P. We know that, radius is perpendicular to tangent at the point of tangency. ⇒CP⊥x+y=0 and CP=r ⇒r=α×1+β×1√(1^2+1^2) ⇒5√2=α+β√2 ⇒50=α+β^22 ⇒α+β^2=100

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

Solution

Relevant Questions

Consider a circle x-α2+y-β2=50, where α,β>0. If the circle touches the line y+x=0 at the point P, whose distance from the origin is 42 , then (α+β)2 is equal to _______.

Solution

Relevant Questions

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