"Let A(1, 4) and B(1, $-$5) be two points. Let P be a point on the circle (x $-$ 1)2 + (y $-$ 1)2 = 1 such that (PA)2 + (PB)2 have maximum value, then the points, P, A and B lie on :"

Question

Accepted Answer

"P be a point on ${(x - 1)^2} + {(y - 1)^2} = 1$

so $P(1 + \cos \theta ,1 + \sin \theta )$

A(1, 4), B(1, $-$5)

${(PA)^2} + {(PB)^2}$

$$ = {(\cos \theta )^2} + {(\sin \theta - 3)^2} + {(\cos \theta )^2} + {(\sin \theta + 6)^2}$$

$= 47 + 6\sin \theta$

It is maximum if $\sin \theta = 1$

When $\sin \theta = 1,\cos \theta = 0$

So P(1, 2), A(1, 4), B(1, $-$5)

P, A, B are collinear points."

Let A(1, 4) and B(1, $-$5) be two points. Let P be a point on the circle
(x $-$ 1)² + (y $-$ 1)² = 1 such that (PA)² + (PB)² have maximum value, then the points, P, A and B lie on :

Solution

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Let A(1, 4) and B(1, $-$5) be two points. Let P be a point on the circle (x $-$ 1)2 + (y $-$ 1)2 = 1 such that (PA)2 + (PB)2 have maximum value, then the points, P, A and B lie on :

Solution

About this question

More Circles questions

Drill 25 more like these. Every day. Free.

Let A(1, 4) and B(1, $-$5) be two points. Let P be a point on the circle
(x $-$ 1)² + (y $-$ 1)² = 1 such that (PA)² + (PB)² have maximum value, then the points, P, A and B lie on :