Given circle is ‌x2+y2−4x+6y+4=0 ‌(x−2)2+(y+3)2=(3)2 ‌‌ Centre ‌=(2,−3)‌ and radius ‌=r=3 ∵ Perpendicular distance between centre and line =r ∴|‌
4×2−3(−3)+p
√42+(3)2
|=3 ⇒‌‌17+p=±15 ‌‌p=−2‌‌(∵p+3>0) So, equation of line is 4x−3y−2=0 Let line touches the circle at (h,k). Then, equation of tangent at (h,k) is ‌hx+ky−2(x+h)+3(y+k)+4=0 ‌(h−2)x+(k+3)y−2h+3k+4=0 On comparing Eq. (ii) with Eq. (i), we get ‌‌
4
h−2
=‌
−3
k+3
‌4k+3h=−6‌‌...‌ (iii) ‌ ‌‌ and ‌‌‌‌
−3
k+3
=‌
−2
−2h+3k+4
‌⇒‌‌7k−6h=−6‌‌...‌ (iv) ‌ On solving Eqs. (iii) and (iv), we get h=‌
