Question

If \( (p, q) \) is the center of the circle which cuts the three circles \( x^2 + y^2 - 2x - 4y + 4 = 0 \), \( x^2 + y^2 + 2x - 4y + 1 = 0 \), and \( x^2 + y^2 - 4x - 2y - 11 = 0 \) orthogonally, then find \( p + q \).

• A. 9
• B. \( \frac{35}{4} \)
• C. \( \frac{15}{2} \)
• D. 7 \bigskip

Hint:

To solve for the center of a circle that cuts three given circles orthogonally, use the power of the point formula and equate it to the square of the radius for each circle. The system of equations will provide the values of \( p \) and \( q \).

Answer 1

The Correct Option is A

We are given three circles:
1) \( x^2 + y^2 - 2x - 4y + 4 = 0 \)
2) \( x^2 + y^2 + 2x - 4y + 1 = 0 \)
3) \( x^2 + y^2 - 4x - 2y - 11 = 0 \)
Let the center of the required circle be \( (p, q) \). This circle cuts the three given circles orthogonally, meaning the power of the center \( (p, q) \) with respect to each of the three circles is equal to the square of the radius of the respective circle. Step 1: Express the power of the point \( (p, q) \) for each circle The power of the point \( (p, q) \) with respect to a circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is given by: \[ \text{Power} = p^2 + q^2 + 2gp + 2fq + c \] Circle 1: \( x^2 + y^2 - 2x - 4y + 4 = 0 \)
Here, \( g = -1 \), \( f = -2 \), and \( c = 4 \).
The power of \( (p, q) \) with respect to this circle is: \[ \text{Power}_1 = p^2 + q^2 - 2p - 4q + 4 \] Circle 2: \( x^2 + y^2 + 2x - 4y + 1 = 0 \)
Here, \( g = 1 \), \( f = -2 \), and \( c = 1 \).
The power of \( (p, q) \) with respect to this circle is: \[ \text{Power}_2 = p^2 + q^2 + 2p - 4q + 1 \] Circle 3: \( x^2 + y^2 - 4x - 2y - 11 = 0 \) Here, \( g = -2 \), \( f = -1 \), and \( c = -11 \). The power of \( (p, q) \) with respect to this circle is: \[ \text{Power}_3 = p^2 + q^2 - 4p - 2q - 11 \] Step 2: Condition for orthogonality Since the circle cuts the three given circles orthogonally, the power of \( (p, q) \) with respect to each circle must be equal to the square of the radius of the respective circle. For circle 1: \[ \text{Radius} = \sqrt{(-1)^2 + (-2)^2 - 4} = \sqrt{1 + 4 - 4} = 1 \] So, the power of \( (p, q) \) with respect to circle 1 is \( 1^2 = 1 \). For circle 2: \[ \text{Radius} = \sqrt{(1)^2 + (-2)^2 - 1} = \sqrt{1 + 4 - 1} = 2 \] So, the power of \( (p, q) \) with respect to circle 2 is \( 2^2 = 4 \). For circle 3: \[ \text{Radius} = \sqrt{(-2)^2 + (-1)^2 + 11} = \sqrt{4 + 1 + 11} = \sqrt{16} = 4 \] So, the power of \( (p, q) \) with respect to circle 3 is \( 4^2 = 16 \). Step 3: Set up equations and solve for \( p \) and \( q \) We have the following system of equations: 1) \( p^2 + q^2 - 2p - 4q + 4 = 1 \)
2) \( p^2 + q^2 + 2p - 4q + 1 = 4 \)
3) \( p^2 + q^2 - 4p - 2q - 11 = 16 \)
Simplify the equations: 1) \( p^2 + q^2 - 2p - 4q = -3 \)
2) \( p^2 + q^2 + 2p - 4q = 3 \)
3) \( p^2 + q^2 - 4p - 2q = 27 \)
Now subtract equations 1 and 2: \[ (p^2 + q^2 + 2p - 4q) - (p^2 + q^2 - 2p - 4q) = 3 - (-3) \] \[ 4p = 6 \] \[ p = \frac{3}{2} \] Next, subtract equations 2 and 3: \[ (p^2 + q^2 - 4p - 2q) - (p^2 + q^2 + 2p - 4q) = 27 - 3 \] \[ -6p + 2q = 24 \] Substitute \( p = \frac{3}{2} \) into the equation: \[ -6 \times \frac{3}{2} + 2q = 24 \] \[ -9 + 2q = 24 \] \[ 2q = 33 \] \[ q = \frac{33}{2} \] Step 4: Find \( p + q \) Now, calculate \( p + q \): \[ p + q = \frac{3}{2} + \frac{33}{2} = \frac{36}{2} = 18 \] Thus, the correct value of \( p + q \) is \( \boxed{9} \). \bigskip