The equation of the circle is ‌x2+y2−6x+2y+c=0 ⇒‌‌(x−3)2+(y+1)2=10−c So, the center of the circle is C(3,−1) and the radius is r2=10−C Now, since the equation of tangent is x−2y=0 Slop, mt=1∕2 Slope of the radius, connecting the centre (3,−1) to the point of tangency P(xp,yp)‌ is ‌mr=‌
yp−(−1)
xp−3
=‌
yp+1
xp−3
Now, since radius is perpendicular to the tangent, so ‌mt⋅mr=−1 ⇒‌
1
2
⋅‌
yp+1
xp−3
=−1⇒yp+1=−2xp+6 ⇒yp=−2xp+5 But, P lies on the tangent line, its coordinates satisfy xp−2yp=0 ⇒‌‌xp=2yp So, ‌‌yp=−2xp+5 ‌⇒‌‌−2(2yp)+5=−4yp+5 ‌⇒‌‌5yp=5⇒yp=1 So, xp=2yp=2×1=2 So, the point P is (2,1). Now, the distance between (6,3) and (2,1) is ‌d=√(2−6)2+(1−3)2 ⇒√(−4)2+(−2)2⇒√16+4=√20=2√5 ∴ The distance of the point (6,3) from P is 2√5.