Question

The radical centre of the system of circles, \[ x^2 + y^2 + 4x + 7 = 0,\quad 2(x^2 + y^2) + 3x + 5y + 9 = 0 \] and \(x^2 + y^2 + y = 0\) is

• A. \((-2,-1)\)
• B. \((1,-2)\)
• C. \((-1,-2)\)
• D. None of these

Hint:

Radical center = solve pairwise subtractions of circle equations.

Answer 1

The Correct Option is A

To find the radical center of the given system of circles, we first need to understand what the radical center is. The radical center is the intersection point of the radical axes of three circles. The radical axis of two circles is the locus of points that have equal power with respect to both circles.

Consider the given circles:

1. Circle 1: \(x^2 + y^2 + 4x + 7 = 0\)
2. Circle 2: \(2(x^2 + y^2) + 3x + 5y + 9 = 0\)
3. Circle 3: \(x^2 + y^2 + y = 0\)

First, convert each circle equation into the standard form:

1. For Circle 1: \((x+2)^2 + y^2 = -4 + 7 = 3\)
2. For Circle 2: Rearrange to \(2x^2 + 2y^2 + 3x + 5y + 9 = 0\). Divide by 2: \(x^2 + y^2 + \frac{3}{2}x + \frac{5}{2}y = -\frac{9}{2}\)
3. For Circle 3: \((x)^2 + (y + \frac{1}{2})^2 = \frac{1}{4}\)

Calculate the radical axis between pairs of circles:

Radical Axis of Circle 1 and Circle 2:

Subtract Circle 1 from Circle 2:

\([(2x^2 + 2y^2 + 3x + 5y + 9) - 2(x^2 + y^2 + 4x + 7)] = (3x + 5y + 9) - (8x + 14)\)

Gives: \(-5x + 5y = -5\), simplify to: \(x - y = -1 \quad \Rightarrow \quad x = y - 1\)

Radical Axis of Circle 1 and Circle 3:

Subtract Circle 1 from Circle 3:

\([x^2 + y^2 + y - (x^2 + y^2 + 4x + 7)] = y - 4x - 7\)

Gives: \(y = 4x + 7\)

The radical center, which lies at the intersection of these two axes, is found by solving:

1. \(x - y = -1\)
2. \(y = 4x + 7\)

Substitute \(y = 4x + 7\) into \(x - y = -1\):

\(x - (4x + 7) = -1 \quad \Rightarrow \quad x - 4x - 7 = -1 \quad \Rightarrow \quad -3x = 6 \quad \Rightarrow \quad x = -2\)

Substitute \(x = -2\) into \(y = 4x + 7\):

\(y = 4(-2) + 7 \quad \Rightarrow \quad y = -8 + 7 = -1\)

Thus, the radical center is \((-2, -1)\).

Therefore, the correct answer is:

Option 1: \((-2, -1)\)