The point where the line 4x-3y+7=0 touches the circle x^2+y^2-6x+4y-12=0 is

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Correct answer: C. (c) Let P (a ,b ) be the point where line 4x-3y+7=0 touches the circle x^2+y^2-6x+4y-12=0 Now, equation of the tangent at P (a ,b ) to the given circle is xa+ yb-3 (x+a )+2 (y+b )-12=0 ⇒ (a-3 )x+ (b+2 )y- (3a-2b+12 )=0 If it represents the given line 4x-3y+7=0 Then, a-34= b+23= - (3a-2b+12 )7=(say) a=4k+3 ,b=-3k-2 and - (3a-2b+12 )=7k ⇒-3a+2b-12=7l … … (i ) Put the value of a and b in eq. (i), -3 (4k+3 )+2 (-3k-2 )-12=7k-12k-9-6k-4-12=7kk=-1a=-1 and b=1

$The point where the line 4 x - 3 y + 7 = 0 touches the circle x^{2} + y^{2} - 6 x + 4 y - 12 = 0 is$

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