If the circle (x-a)2+y2=25 intersects the circle x2+(y-b)2=16in such a way that common chord is of maximum length, then value of a2+b2is
Answers
Given : circles (x - a)² + y² = 25 and x² + (y-b)² = 16 intersects in such a way that common chord is of maximum length,
To Find : Value of a² + b²
Solution:
(x - a)² + y² = 25 => (x - a)² + y² = 5² center (a , 0) , Radius = 5
x² + (y-b)² = 16 => x² + (y-b)² = 4² center (0 , b) , Radius = 4
Common Chord can be maximum equal to the Diamter of smaller circle
Smaller circle is x² + (y-b)² = 4²
Diamter = 2 x 4 = 8
Perpendicular bisector of chord will pass through center of other circle
Radius of other Circle = 5
Half of chord = diameter/2 = radius = 4
Distance between center of circles
=√( 5² - 4²)
=√ 25 -16
= √9
= 3
center (a , 0) and center (0 , b)
Distance between centers = √a² + b²
√a² + b² = 3
=> a² + b² = 9
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