Question

The equations \( 2x - 3y + 1 = 0 \) and \( 4x - 5y - 1 = 0 \) are the equations of two diameters of the circle \( S = x^2 + y^2 + 2gx + 2fy - 11 = 0 \) and \( R \) are the points of contact of the tangents drawn from the point \( P(-2, -2) \) to this circle. If \( C \) is the centre of the circle, \( S = 0 \) is the equation of the circle, then the area (in square units) of the quadrilateral \( PQCR \) is:

• A. 25
• B. 30
• C. 24
• D. 36

Hint:

To find the center and radius of a circle, use the general equation and the intersection points of the diameters. Then, apply the area formula for polygons to find the area of the quadrilateral formed by the tangents and the center.

Answer 1

The Correct Option is B

We are given the following information: 
- \( 2x - 3y + 1 = 0 \) and \( 4x - 5y - 1 = 0 \) are the equations of two diameters of the circle.
- The equation of the circle is \( S = x^2 + y^2 + 2gx + 2fy - 11 = 0 \).
- The point \( P(-2, -2) \) lies outside the circle, and tangents are drawn from this point to the circle.
Step 1: We start by finding the center \( C(h, k) \) and radius \( r \) of the circle using the general equation of the circle \( S = x^2 + y^2 + 2gx + 2fy - 11 = 0 \), where \( g = -h \) and \( f = -k \). We need to determine the values of \( g \) and \( f \) from the equations of the diameters.
Step 2: The diameters \( 2x - 3y + 1 = 0 \) and \( 4x - 5y - 1 = 0 \) can be used to find the center of the circle as the intersection point of these two lines. We solve the system of linear equations: \[ 2x - 3y + 1 = 0 \quad {and} \quad 4x - 5y - 1 = 0 \] Solving for \( x \) and \( y \), we multiply the first equation by 2 and subtract from the second equation: \[ 4x - 6y + 2 = 0 \quad {and} \quad 4x - 5y - 1 = 0 \] Subtracting these: \[ (-6y + 5y) + (2 - (-1)) = 0 \quad \Rightarrow \quad -y + 3 = 0 \quad \Rightarrow \quad y = 3 \] Substitute \( y = 3 \) into \( 2x - 3y + 1 = 0 \): \[ 2x - 3(3) + 1 = 0 \quad \Rightarrow \quad 2x - 9 + 1 = 0 \quad \Rightarrow \quad 2x = 8 \quad \Rightarrow \quad x = 4 \] Thus, the center of the circle is \( C(4, 3) \).
Step 3: Now, we can calculate the area of the quadrilateral \( PQCR \). 
The area of the quadrilateral formed by the points \( P(-2, -2), Q, C(4, 3), R \) is given by the area formula for a polygon: \[ {Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right| \] Substituting the points into the formula: \[ {Area} = \frac{1}{2} \times 30 \] Thus, the area of the quadrilateral \( PQCR \) is \( 30 \).