Question

If the point of contact of the circles \( x^2+y^2-6x-4y+9=0 \) and \( x^2+y^2+2x+2y-7=0 \) is \( (\alpha, \beta) \), then \( 7\beta = \)

• A. \( 5\alpha \)
• B. \( 2\alpha \)
• C. \( 3\alpha \)
• D. \( 4\alpha \)

Hint:

1. Find centres \(C_1, C_2\) and radii \(R_1, R_2\). 2. Check for tangency: if distance \(C_1C_2 = R_1+R_2\), they touch externally. If \(C_1C_2 = |R_1-R_2|\), they touch internally. 3. For external tangency, the point of contact P divides the line segment \(C_1C_2\) internally in the ratio \(R_1:R_2\). The point P can be found using the section formula as \(P = \frac{R_2 C_1 + R_1 C_2}{R_1+R_2}\).

Answer 1

The Correct Option is D

Circle \(S_1: x^2+y^2-6x-4y+9=0 \).
Centre \( C_1 = (3,2) \).
Radius \( R_1 = \sqrt{3^2+2^2-9} = \sqrt{9+4-9} = \sqrt{4} = 2 \).
Circle \(S_2: x^2+y^2+2x+2y-7=0 \).
Centre \( C_2 = (-1,-1) \).
Radius \( R_2 = \sqrt{(-1)^2+(-1)^2-(-7)} = \sqrt{1+1+7} = \sqrt{9} = 3 \).
Distance between centres \( C_1C_2 = \sqrt{(3-(-1))^2 + (2-(-1))^2} = \sqrt{4^2+3^2} = \sqrt{16+9} = \sqrt{25} = 5 \).
Sum of radii \( R_1+R_2 = 2+3 = 5 \).
Since \( C_1C_2 = R_1+R_2 \), the circles touch externally.
The point of contact P\((\alpha, \beta)\) divides the line segment \(C_1C_2\) internally in the ratio \(R_1:R_2 = 2:3\).
Let \(C_1=(x_1,y_1)=(3,2)\) and \(C_2=(x_2,y_2)=(-1,-1)\).
The point of contact P divides \(C_1C_2\) in ratio \(R_1:R_2\).
The formula for the point P that divides segment \(C_1C_2\) in ratio \(m:n\) is \( P = \left( \frac{nx_1+mx_2}{m+n}, \frac{ny_1+my_2}{m+n} \right) \).
Here \(m=R_1=2, n=R_2=3\).
So, \( P = \left( \frac{R_2 x_1 + R_1 x_2}{R_1+R_2}, \frac{R_2 y_1 + R_1 y_2}{R_1+R_2} \right) \).
\[ \alpha = \frac{3(3) + 2(-1)}{2+3} = \frac{9-2}{5} = \frac{7}{5} \] \[ \beta = \frac{3(2) + 2(-1)}{2+3} = \frac{6-2}{5} = \frac{4}{5} \] Point of contact \( (\alpha, \beta) = \left(\frac{7}{5}, \frac{4}{5}\right) \).
We need to check the relation for \( 7\beta \): \[ 7\beta = 7 \times \frac{4}{5} = \frac{28}{5} \] Compare with options based on \( \alpha = 7/5 \): Option (4): \( 4\alpha = 4 \times \frac{7}{5} = \frac{28}{5} \).
So, \( 7\beta = 4\alpha \).
This matches option (4).