JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
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 \(24 A\) is equal to ...... .
- A \(1140\)
- B \(1225\)
- C \(2450\)
- D \(612\)
Answer & Solution
Correct Answer
(B) \(1225\)
Step-by-step Solution
Detailed explanation
Equation of normal \(4 x-3 y+1=0\) and equation of tangents \(3 x+4 y-43=0\) Area of triangle \(=\frac{1}{2}\left(\frac{43}{3}+\frac{1}{4}\right) \times(7)\) \(=\frac{1}{2}\left(\frac{172+3}{12}\right) \times 7\) \(A =\frac{1225}{24}\) \(24 A =1225\)
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