If the centre $(\alpha, \beta)$ of a circle cutting the circles $x^2+y^2-2y-3=0$ and $x^2+y^2+4x+3=0$ orthogonally lies on the line $2x-3y+4=0$, then $2\alpha+\beta=$
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The locus of the center of a circle that cuts two given circles orthogonally is their radical axis. This is a very useful property that can simplify problems significantly. Finding the radical axis by simply subtracting the equations of the circles (in the standard form with leading coefficients of 1) is a quick and effective method.
Step 1: Radical Axis:
The locus of centers of circles cutting two circles orthogonally is their radical axis.
The equation of the required circle can be found, but simply, the center \( (\alpha, \beta) \) must lie on the radical axis of the two given circles.
Radical Axis \( S_1 - S_2 = 0 \):
\( (x^2+y^2-2y-3) - (x^2+y^2+4x+3) = 0 \)
\( -4x - 2y - 6 = 0 \)
\( 2x + y + 3 = 0 \).
So, \( 2\alpha + \beta + 3 = 0 \).
Step 2: Calculate Value:
\( 2\alpha + \beta = -3 \).
(The additional information about the line \( 2x-3y+4=0 \) is usually to fix the center uniquely, but the value required depends only on the radical axis relation).