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 \_\_\_\_\_\_\_\_.
Answer: 1225
Solutions · 1
Official solution
*(This question, though listed under Applications of Integrals in the source, is a coordinate-geometry tangent/normal problem, not an area-under-curve integral computation. Center (2,3), point P(5,7).)*
Normal at P (through center): slope =5−27−3=34, giving line 4x−3y+1=0, meeting the x-axis at M(−41,0).
Tangent at P has slope −43: 3x+4y=43, meeting the x-axis at N(343,0).
\[A = \frac12\times MN\times PQ = \frac12\times\frac{175}{12}\times7 = \frac{1225}{24}\]
24A=1225
(Note: the source flags this JEE question as technically inconsistent, since the triangle as described does not actually lie entirely along the positive x-axis; the numerical answer 1225 is the official key value.)