Test Index

Circles Part 1

Section: Mathematics

© examsnet.com

Question : 25 of 100

Marks: +1, -0

Let

A(1,4)

B(1,-5)

be two points. Let

P

be a point on the circle

(x-1)^{2}+(y-1)^{2}=1

(P A)^{2}+(P B)^{2}

have maximum value, then the points

P, A

B

[26 Feb 2021 Shift 2]

a straight line
a hyperbola
an ellipse
a parabola

Solution:

P

be a point on circle

(x-1)^{2}+(y-1)^{2}=1

So, coordinate of

P

will be of form

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

Given, points

A(1,4)

and

B(1,-5)

Then,

\;(P A)^{2}+(P B)^{2}

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

\;\; =2 \cos ^{2} \theta+2 \sin ^{2} \theta-6 \sin \theta+12 \sin \theta+45

\;\; =47+6 \sin \theta

Now,

(P A)^{2}+(P B)^{2}=47+6 \sin \theta \; \text{ is } \;

Maximum, when

\sin \theta=1

\theta\; \theta\; =\frac{\pi}{2}

\Rightarrow\; \cos \theta\; =0

\;P(1,2), A(1,4), B(1,-5)

\therefore P, A, B

are collinear points on the line

x=1

© examsnet.com

Go to Question:

Prev Question

Next Question