JEE PYQ: Circle Question 11
Question 11 - 2021 (24 Feb Shift 2)
If the area of the triangle formed by the positive $x$-axis, the normal and the tangent to the circle $(x-2)^2 + (y-3)^2 = 25$ at the point $(5, 7)$ is $A$, then $24A$ is equal to
Note: NTA has dropped this question in the final official answer key.
Show Answer
Answer: 1225
Solution
Normal at $P(5,7)$ passes through center $(2,3)$: slope $= \frac{4}{3}$. Normal meets $x$-axis at $M(-\frac{1}{4}, 0)$. Tangent: slope $= -\frac{3}{4}$, meets $x$-axis at $N(\frac{43}{3}, 0)$. $24A = 24 \cdot \frac{1}{2} \cdot MN \cdot 7 = 1225$.
BETA
