Find at what points on the circle x^2 + y^2 = 13, the tangent is parallel to the line 2x + 3y = 7
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The line parallel to 2x+3y-7 = 0 is of the form 2x+3y+k = 0.
The perpendicular distance from centre to the tangent is equal to radius of the circle.
Center of given circle is (0,0)
Radius = √13
The perpendicular distance = [|2(0)+3(0)+k|]/√4+9
But we know perpendicular distance = radius
|k|/√13 = √13
|k|= 13.
k = + 13
The equation of tangents are
2x+3y+13 = 0 and 2x+3y-13 = 0
The perpendicular distance from centre to the tangent is equal to radius of the circle.
Center of given circle is (0,0)
Radius = √13
The perpendicular distance = [|2(0)+3(0)+k|]/√4+9
But we know perpendicular distance = radius
|k|/√13 = √13
|k|= 13.
k = + 13
The equation of tangents are
2x+3y+13 = 0 and 2x+3y-13 = 0
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