JEE Main Math Class 12 Application of Integrals Questions

Given, circle (x−2)2+(y−3)2=52
‌c=(2,3)
‌r=5 Equation of normal at P (i.e. PA line)
⇒‌‌(y−7)=(‌

7−3

5−2

)(x−5)
⇒‌‌3y−21=4x−20
⇒‌‌4x−3y+1=0
Therefore, M=(‌

−1

4

,0) [Put y=0 in above equation]
Now, equation of tangent at P.
y−7‌‌=‌

−3

4

(x−5)‌‌[∵‌ slope of ‌PN=‌

−1

‌ Slope of ‌PA

]
⇒4y−28‌‌=−3x+15
⇒3x+4y‌‌=43
Therefore, N=(‌

43

3

,0) [Put y=0 in above equation]
∴‌ Area ‌(A)‌‌=‌

1

2

×MN×PQ
‌‌=‌

1

2

×(‌

43

3

+‌

1

4

)×7
‌‌=‌

1

2

×‌

175

12

×7
∴‌‌24A‌‌=24×‌

1

2

×‌

175

12

×7=1225
But this question is wrong as in question. It is mentioned that the triangle is formed with the positive X-axis which contradicts the solution.