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

If $P(2,3)$ and $Q(-1,2)$ are conjugate points with respect to the circle $x^2+y^2+2gx+3y-2=0$,then the radius of the circle is

$A$ circular wire of radius $7\,cm$ is cut and bent again into an arc of a circle of radius $12\,cm$. The angle subtended by the arc at the centre is ......$^o$

The power of a point $(2,0)$ with respect to a circle $S$ is $-4$ and the length of the tangent drawn from the point $(1,1)$ to $S$ is $2$. If the circle $S$ passes through the point $(-1,-1)$,then the radius of the circle $S$ is

Find the equation of the chord of the circle $x^2 + y^2 = 16$ which is bisected at the point $(2, -1)$.

If $y=2x$ is a chord of the circle $x^2+y^2-10x=0$,then the equation of the circle with this chord as diameter is

Vedclass Test Series

Exam Paper Generator

Online Exam Module