Let A(1, 4) and B(1, -5) be two points. Let P | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $P$ be a point on the circle $(x-1)^{2} + (y-1)^{2} = 1$.We can parameterize $P$ as $(1 + \cos 	heta, 1 + \sin 	heta)$.Given $A(1, 4)$ and $

Vedclass · Accepted Answer

a straight line

Vedclass · Answer

(A) Let $P$ be a point on the circle $(x-1)^{2} + (y-1)^{2} = 1$.We can parameterize $P$ as $(1 + \cos 	heta, 1 + \sin 	heta)$.Given $A(1, 4)$ and $B(1, -5)$,we calculate $(PA)^{2} + (PB)^{2}$:$(PA)^{2} = (1 + \cos 	heta - 1)^{2} + (1 + \sin 	heta - 4)^{2} = \cos^{2} 	heta + (\sin 	heta - 3)^{2} = \cos^{2} 	heta + \sin^{2} 	heta - 6 \sin 	heta + 9 = 10 - 6 \sin 	heta$.$(PB)^{2} = (1 + \cos 	heta - 1)^{2} + (1 + \sin 	heta + 5)^{2} = \cos^{2} 	heta + (\sin 	heta + 6)^{2} = \cos^{2} 	heta + \sin^{2} 	heta + 12 \sin 	heta + 36 = 37 + 12 \sin 	heta$.Summing these,$(PA)^{2} + (PB)^{2} = 10 - 6 \sin 	heta + 37 + 12 \sin 	heta = 47 + 6 \sin 	heta$.This expression is maximized when $\sin 	heta = 1$.For $\sin 	heta = 1$,we have $\cos 	heta = 0$,so $P = (1 + 0, 1 + 1) = (1, 2)$.The points are $P(1, 2)$,$A(1, 4)$,and $B(1, -5)$.Since all points have an $x$-coordinate of $1$,they all lie on the vertical line $x = 1$,which is a straight line.

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 $(PA)^{2} + (PB)^{2}$ has a maximum value. Then the points $P, A,$ and $B$ lie on:

Similar Questions

The angle between two circles,each passing through the centre of the other,is

Let the normals at all the points on a given curve pass through a fixed point $(a, b)$. If the curve passes through $(3, -3)$ and $(4, -2\sqrt{2})$,and given that $a - 2\sqrt{2}b = 3$,then $(a^{2} + b^{2} + ab)$ is equal to ..... .

For the line $3x + 2y = 12$ and the circle $x^2 + y^2 - 4x - 6y + 3 = 0$,which of the following statements is true?

If the straight lines $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$ are the tangents to the same circle,then the area of that circle (in square units) is

The equation of the incircle of the triangle formed by the lines $x=0, y=0$ and $3x+4y-24=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module