Find the equation of the circle passing through the point \( (3,4) \) and having equations of two of its diameters \( x+y-14=0 \) and \( 2 x-y-4=0 \quad \) Ans: \( x^{2}+y^{2}-12 x-16 y+75=0 \)

To find the equation of the circle that passes through the point \( (3,4) \) and has diameters defined by the lines \( x+y-14=0 \) and \( 2x-y-4=0 \), we can follow these steps: 1. **Find the intersection of the diameters**: The center of the circle will be at the intersection of the two diameters. 2. **Calculate the intersection point**: We will solve the system of equations given by the two lines. 3. **Use the center and the point on the circle to find the radius**: The radius can be calculated as the distance from the center to the point \( (3,4) \). 4. **Formulate the equation of the circle**: Using the center and radius, we can write the equation of the circle. ### Step 1: Find the intersection of the diameters We have the equations: 1. \( x + y - 14 = 0 \) (Equation 1) 2. \( 2x - y - 4 = 0 \) (Equation 2) Let's solve these equations simultaneously. ### Step 2: Solve the system of equations From Equation 1, we can express \( y \) in terms of \( x \): \[ y = 14 - x \] Now, substitute \( y \) into Equation 2: \[ 2x - (14 - x) - 4 = 0 \] \[ 2x - 14 + x - 4 = 0 \] \[ 3x - 18 = 0 \] \[ x = 6 \] Now substitute \( x = 6 \) back into the expression for \( y \): \[ y = 14 - 6 = 8 \] Thus, the center of the circle is at the point \( (6, 8) \). ### Step 3: Calculate the radius Now we need to find the radius, which is the distance from the center \( (6, 8) \) to the point \( (3, 4) \): \[ r = \sqrt{(6 - 3)^2 + (8 - 4)^2} \] \[ r = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \] ### Step 4: Formulate the equation of the circle The general equation of a circle with center \( (h, k) \) and radius \( r \) is given by: \[ (x - h)^2 + (y - k)^2 = r^2 \] Substituting \( h = 6 \), \( k = 8 \), and \( r = 5 \): \[ (x - 6)^2 + (y - 8)^2 = 5^2 \] \[ (x - 6)^2 + (y - 8)^2 = 25 \] Expanding this equation: \[ (x^2 - 12x + 36) + (y^2 - 16y + 64) = 25 \] \[ x^2 + y^2 - 12x - 16y + 100 = 25 \] \[ x^2 + y^2 - 12x - 16y + 75 = 0 \] Thus, the equation of the circle is: \[ \boxed{x^2 + y^2 - 12x - 16y + 75 = 0} \]