Let point of contact be (x1,y1). Given, equation of circle is x2+y2−4x−6y−12=0 ∴ Equation of chord of contact at point (x1,y1) is xx1+yy1−2(x+x1)−3(y+y1)−12‌=0 ⇒‌‌x(x1−2)+y(y1−3)−2x1−3y1−12‌=0 But it is given that point of contact is ‌3x+4y+7=0 ‌∴‌‌‌
x1−2
3
=‌
y1−3
4
=−‌
(2x1+3y1+12)
7
On taking Ist and last terms, 7x1−14=−(6x1+9y1+36) ⇒‌‌13x1+9y1+22=0 On taking IInd and last terms, ‌7y1−21=−(8x1+12y1+48) ⇒‌‌8x1+19y1+27=0 On solving Eqs. (i) and (ii), we get x1=−1,y=−1 Hence, required point of contact is (−1,−1).