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Q.
If the equation of the tangent to the circle $ {{x}^{2}}+{{y}^{2}}-2x+6y-6=0 $ parallel to $ 3x-4y+7=0 $ is $ 3x-4y+k=0, $ then the values of $ k $ are:
The given equation of circle is $ {{x}^{2}}+{{y}^{2}}-2x+6y-6=0 $ The centre and radius of circle are $ (1,-3) $ and 4 respectively. Length of perpendicular from $ (1,-3) $ to $ 3x-4y+k=0 $ is equal to radius 4. $ \therefore $ $ \left| \frac{3+12+k}{\sqrt{9+16}} \right|=4 $ $ \Rightarrow $ $ 15+k=\pm 20 $ $ \Rightarrow $ $ k=20-15=5 $ Or $ k=-20-15=-35 $