Find the equations of those tangents to the circle x^2 + y^2 = 52, which are parallel to the straight line 2x + 3y = 6.
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let the equation of tangents is 2x + 3y = c
then perpendicular from center of the circle = radius
center = (0,0)
|{(0*2 + 0*3 - c)/(2^2 + 3^2)^1/2}| = 52^1/2
|c/13^1/2| = 52^1/2
c = 2*13 or c = -2*13
c = 26 or c = -26 hence equation of tangents
2x+3y = 26 & 2x+3y = -26
then perpendicular from center of the circle = radius
center = (0,0)
|{(0*2 + 0*3 - c)/(2^2 + 3^2)^1/2}| = 52^1/2
|c/13^1/2| = 52^1/2
c = 2*13 or c = -2*13
c = 26 or c = -26 hence equation of tangents
2x+3y = 26 & 2x+3y = -26
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