Question

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 _______

Hint:

The normal to a circle is simply the line passing through the point of tangency and the center.

Answer 1

Correct Answer: 1225

Step 1: Center $C(2, 3)$, Point $P(5, 7)$. Slope $m_{radius} = \frac{7-3}{5-2} = \frac{4}{3}$.
Step 2: Tangent slope $m_T = -3/4$. Eq: $y-7 = -\frac{3}{4}(x-5) \Rightarrow 3x + 4y = 43$. X-intercept ($y=0$): $x_T = 43/3$.
Step 3: Normal is the line $CP$. Eq: $y-7 = \frac{4}{3}(x-5) \Rightarrow 4x - 3y = -1$. X-intercept ($y=0$): $x_N = -1/4$.
Step 4: Base of triangle on x-axis: $|x_T - x_N| = |\frac{43}{3} + \frac{1}{4}| = \frac{175}{12}$.
Step 5: Height of triangle = y-coordinate of $P = 7$.
Step 6: $A = \frac{1}{2} \times \frac{175}{12} \times 7 = \frac{1225}{24} \Rightarrow 24A = 1225$.