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# The coordinates of any point on the circle through the points $A\left( 2,2 \right)$, $B\left( 5,3 \right)$ and $C\left( 3,-1 \right)$ can be written in the form $\left( 4+\sqrt{5}\cos \theta ,1+\sqrt{5}\sin \theta \right)$. Then the coordinates of the point$P$ on $BC$ such that $AP$ is perpendicular to $BC$ are (a)$\left( -1,4 \right)$ (b)$\left( 4,1 \right)$ (c) $\left( 1,4 \right)$ (d)$\left( 2,3 \right)$

Last updated date: 24th Jul 2024
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Hint: To find the coordinates of point $P$ find the equation of line joining any two points and assume any point $P$ on the line. Use the fact that the product of slopes of two perpendicular lines is $-1$.

We have the points $A\left( 2,2 \right)$, $B\left( 5,3 \right)$ and $C(3,-1)$ on the circle. Other points on the circle are of the form $\left( 4+\sqrt{5}\cos \theta ,1+\sqrt{5}\sin \theta \right)$. We want to find the coordinates of point $P$ such that the line $AP$ is perpendicular to the line $BC$.
We will begin by finding the equation of line $BC$.
We know that the equation of line joining any two points $(a,b)$ and $\left( c,d \right)$ is $\left( y-b \right)=\dfrac{d-b}{c-a}\left( x-a \right)$.
To find the equation of line $BC$, we will substitute $a=5,b=3,c=3,d=-1$ in the above formula.
Thus, we have $y-3=\dfrac{-1-3}{3-5}\left( x-5 \right)$ as the equation of line $BC$.
\begin{align} & \Rightarrow y-3=2\left( x-5 \right) \\ & \Rightarrow y-3=2x-10 \\ & \Rightarrow y=2x-7 \\ \end{align}
Hence, the equation of line $BC$ is $y=2x-7$. $...\left( 1 \right)$
Now, we will find the point $P$.
Let’s assume $P$ has coordinates $(u,v)$.
Substituting $(u,v)$ in the equation of line, we have $v=2u-7$. $...\left( 2 \right)$
To find the slope of line $AP$, we will substitute $a=2,b=2,c=u,d=2u-7$ in the formula $\left( \dfrac{d-b}{c-a} \right)$ of the slope of line.
Thus, we have $\dfrac{2u-7-2}{u-2}=\dfrac{2u-9}{u-2}$ as the slope of $AP$.
We know that the product of slopes of two perpendicular lines is $-1$.
Thus, as $AP$ and $BC$ are perpendicular to each other, the product of their slopes is $-1$.
Using equation $\left( 1 \right)$ and $\left( 2 \right)$, we have $2\left( \dfrac{2u-9}{u-2} \right)=-1$ as slope of line $BC$ is $-1$.
On further solving the equation, we have $4u-18=-u+2$.
\begin{align} & \Rightarrow 5u=20 \\ & \Rightarrow u=4 \\ \end{align}
Substituting the above value in equation $\left( 2 \right)$, we get $v=2u-7=2\left( 4 \right)-7=8-7=1$.
Thus, the coordinates of $P$ are $(u,v)=\left( 4,1 \right)$.
Hence, the correct answer is $\left( 4,1 \right)$.
So, the answer is Option (b)

Note: One must observe that the point $P$ is on the line $BC$ and not on the circle. If you take point $P$ on the circle, you will get an incorrect answer.