Question

If P(\(\alpha, \beta\)) is the radical centre of the circles \(S=x^2+y^2+4x+7=0\), \(S'=2x^2+2y^2+3x+5y+9=0\) and \(S''=x^2+y^2+y=0\), then the length of the tangent drawn from P to S' = 0 is

• A. \(5\)
• B. \(8\)
• C. \(4\)
• D. \(2\)

Hint:

To find the radical centre of three circles, first normalize all circle equations so that the coefficients of \(x^2\) and \(y^2\) are 1. Then, find the equations of at least two radical axes by subtracting the equations of the circles pairwise (\(S_1-S_2=0\), \(S_2-S_3=0\), etc.). The intersection point of these radical axes is the radical centre. The length of the tangent from a point \((x_1, y_1)\) to a circle \(S=0\) is \(\sqrt{S(x_1, y_1)}\), where \(S(x_1, y_1)\) is the result of substituting the point's coordinates into the normalized circle equation.

Answer 1

The Correct Option is D

Step 1: Normalize the equations of the circles if necessary.
The equations of the given circles are:
\(S: x^2 + y^2 + 4x + 7 = 0\)
\(S': 2x^2 + 2y^2 + 3x + 5y + 9 = 0\)
\(S'': x^2 + y^2 + y = 0\)
For finding radical axes, the coefficients of \(x^2\) and \(y^2\) in all circle equations must be 1. Normalize \(S'\) by dividing by 2:
\(S': x^2 + y^2 + \frac{3}{2}x + \frac{5}{2}y + \frac{9}{2} = 0\).
Step 2: Find the equations of any two radical axes.
The radical axis of two circles \(S_1 = 0\) and \(S_2 = 0\) is given by \(S_1 - S_2 = 0\).
Radical Axis of \(S\) and \(S'\) (\(RA_{S,S'}\)):
\(RA_{S,S'} = S - S' = (x^2 + y^2 + 4x + 7) - (x^2 + y^2 + \frac{3}{2}x + \frac{5}{2}y + \frac{9}{2}) = 0\)
\[ \left(4 - \frac{3}{2}\right)x - \frac{5}{2}y + \left(7 - \frac{9}{2}\right) = 0 \] \[ \frac{8 - 3}{2}x - \frac{5}{2}y + \frac{14 - 9}{2} = 0 \] \[ \frac{5}{2}x - \frac{5}{2}y + \frac{5}{2} = 0 \] Multiply by \(\frac{2}{5}\):
\[ x - y + 1 = 0 \quad \text{(Equation 1)} \] Radical Axis of \(S\) and \(S''\) (\(RA_{S,S''}\)):
\(RA_{S,S''} = S - S'' = (x^2 + y^2 + 4x + 7) - (x^2 + y^2 + y) = 0\)
\[ 4x - y + 7 = 0 \quad \text{(Equation 2)} \] Step 3: Find the coordinates of the radical centre P(\(\alpha, \beta\)).
The radical centre is the point of intersection of the radical axes.
Solve Equation 1 and Equation 2 simultaneously.
From Equation 1, we can express \(y\) in terms of \(x\):
\(y = x + 1\)
Substitute this into Equation 2:
\(4x - (x + 1) + 7 = 0\)
\(4x - x - 1 + 7 = 0\)
\(3x + 6 = 0\)
\(3x = -6\)
\(x = -2\)
Now substitute \(x = -2\) back into \(y = x + 1\):
\(y = -2 + 1 = -1\)
So, the radical centre P is \((\alpha, \beta) = (-2, -1)\).
Step 4: Calculate the length of the tangent drawn from P to \(S' = 0\).
The length of the tangent from a point \((x_1, y_1)\) to a circle \(x^2 + y^2 + 2gx + 2fy + c = 0\) is given by \(\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}\). This is also denoted as \(\sqrt{S_1}\).
The point P is \((\alpha, \beta) = (-2, -1)\).
The circle is the normalized \(S': x^2 + y^2 + \frac{3}{2}x + \frac{5}{2}y + \frac{9}{2} = 0\).
Length of tangent \(L = \sqrt{S'(\alpha, \beta)}\)
\[ L = \sqrt{(-2)^2 + (-1)^2 + \frac{3}{2}(-2) + \frac{5}{2}(-1) + \frac{9}{2}} \] \[ L = \sqrt{4 + 1 - 3 - \frac{5}{2} + \frac{9}{2}} \] \[ L = \sqrt{2 + \frac{9 - 5}{2}} \] \[ L = \sqrt{2 + \frac{4}{2}} \] \[ L = \sqrt{2 + 2} \] \[ L = \sqrt{4} \] \[ L = 2. \]