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Mathematics · 2024 · 2 marks
CBSE 2024 · Region 3 · Set 1 · Q23
Find a relation between $\displaystyle x$ and y such that the point $\displaystyle \mathrm{P}(x, \mathrm{y})$ is equidistant from the points A$\displaystyle (7, 1)$ and B$\displaystyle (3, 5)$.Points A(-$\displaystyle 1$, y) and B$\displaystyle (5, 7)$ lie on a circle with centre O($\displaystyle 2$, -3y) such that AB is a diameter of the circle. Find the value of y. Also, find the radius of the circle.
Find a relation between $\displaystyle x$ and y such that the point $\displaystyle \mathrm{P}(x, \mathrm{y})$ is equidistant from the points A$\displaystyle (7, 1)$ and B$\displaystyle (3, 5)$.
Points A(-$\displaystyle 1$, y) and B$\displaystyle (5, 7)$ lie on a circle with centre O($\displaystyle 2$, -3y) such that AB is a diameter of the circle. Find the value of y. Also, find the radius of the circle.
Marking-scheme solution
(a)
\[\begin{aligned}
& \mathrm{PA}=\mathrm{PB} \\
& \Rightarrow \mathrm{PA}^{2}=\mathrm{PB}^{2} \\
&(\mathrm{x}-7)^{2}+(\mathrm{y}-1)^{2}=(\mathrm{x}-3)^{2}+(\mathrm{y}-5)^{2} \\
& \Rightarrow-8 \mathrm{x}+8 \mathrm{y}+16=0 \text { or } \mathrm{x}-\mathrm{y}-2=0
\end{aligned}
\]
(b)
Centre O ($\displaystyle 2$, - 3y) is the mid point of AB
\[\begin{aligned}
& \therefore \frac{y+7}{2}=-3 y \\
& \Rightarrow y=-1
\end{aligned}
\]
Radius \(\displaystyle =\mathrm{OB}=\sqrt{(5-2)^{2}+(7-3)^{2}}=5\)
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CBSE Class 10 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.