Let the equation of the circle, which touches x -axis at the point ( a , 0 ) , a 0 and cuts off an intercept of length b on y -axis be x 2 + y 2 − αx + β y + γ = 0 . If the circle lies below x -axis

Question

Let the equation of the circle, which touches x -axis at the point ( a , 0 ) , a > 0 and cuts off an intercept of length b on y -axis be x 2 + y 2 − αx + β y + γ = 0 . If the circle lies below x -axis, then the ordered pair ( 2 a , b 2 ) is equal to

TestHub · Accepted Answer

By pythagoras r 2 = a 2 + 4 b 2 ​ = P 2 r = 4 4 a 2 + b 2 ​ ​ Equation of circle is ( x − α ) 2 + ( y − β ) 2 = r 2 x 2 + y 2 − 2 a x − 2 p y + α 2 + p 2 − r 2 = 0 comparision x 2 + y 2 − αx + β y + r = 0 − α = − 2 a , β = − 2 p , r = a 2 ⇒ 2 a = α , 4 a 2 + b 2 = 4 p 2 α 2 + b 2 = 4 p 2 α 2 + b 2 = β 2 ​ So, ( 2 a , b 2 ) = ( α , β 2 − 4 r )

Mathematics - Circle Question with Solution | TestHub

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