Question

The tangents drawn from a point (2,-1) touch the circle \(x^2+y^2+4x-2y+1=0\) at the points A and B. If C is the centre of the circle, then the area (in sq. units) of the triangle ABC is

• A. \( \frac{4}{5} \)
• B. 4
• C. 8
• D. \( \frac{8}{5} \)

Hint:

A useful formula for the area of the triangle formed by the points of contact (A, B) and the external point (P) is \( \text{Area} = \frac{r L^3}{r^2 + L^2} \). The area of the triangle formed by the points of contact and the center (C) is \( \text{Area} = \frac{r^3 L}{r^2 + L^2} \). In this case, \( \frac{2^3 \cdot 4}{2^2+4^2} = \frac{8 \cdot 4}{4+16} = \frac{32}{20} = \frac{8}{5} \).

Answer 1

The Correct Option is D

Let the point be P(2,-1).
Step 1: Find the center and radius of the circle.
Circle: \(x^2+y^2+4x-2y+1=0\).
Center \(C = (-2, 1)\).
Radius \(r = \sqrt{(-2)^2+1^2-1} = \sqrt{4+1-1} = \sqrt{4} = 2\).
Step 2: Find the length of the tangent from P to the circle.
The length of the tangent, \(L\), is given by \( \sqrt{S_{11}} \).
\( L = \sqrt{(2)^2 + (-1)^2 + 4(2) - 2(-1) + 1} = \sqrt{4+1+8+2+1} = \sqrt{16} = 4 \).
Step 3: Use the formula for the area of the triangle formed by the points of contact and the center.
A convenient formula for the area of triangle ABC is \( \text{Area} = \frac{rL^3}{r^2+L^2} \). Wait, this is for triangle PAB.
Let's use a more fundamental method. The area of the kite PACB is Area = \( r \times L = 2 \times 4 = 8 \).
Let \(d\) be the distance PC. \(d = \sqrt{(2 - (-2))^2 + (-1 - 1)^2} = \sqrt{4^2 + (-2)^2} = \sqrt{16+4} = \sqrt{20}\).
The area of \(\triangle PAC = \frac{1}{2} r L = \frac{1}{2}(2)(4) = 4\).
Let \(h\) be the altitude from A to PC in \(\triangle PAC\). This \(h\) is half the length of the chord of contact AB.
Area(\(\triangle PAC\)) = \( \frac{1}{2} \times PC \times h = \frac{1}{2} d h \). So, \( 4 = \frac{1}{2} \sqrt{20} h \implies h = \frac{8}{\sqrt{20}} = \frac{4}{\sqrt{5}} \).
The length of the chord AB is \(2h = \frac{8}{\sqrt{5}}\).
Let M be the intersection of AB and PC. In \(\triangle CMA\), \(CM^2 = CA^2 - AM^2 = r^2 - h^2 = 2^2 - (\frac{4}{\sqrt{5}})^2 = 4 - \frac{16}{5} = \frac{4}{5}\). So \(CM = \frac{2}{\sqrt{5}}\).
The area of triangle ABC is \( \frac{1}{2} \times \text{base AB} \times \text{height CM} \).
Area(ABC) = \( \frac{1}{2} \times \left(\frac{8}{\sqrt{5}}\right) \times \left(\frac{2}{\sqrt{5}}\right) = \frac{1}{2} \times \frac{16}{5} = \frac{8}{5} \).