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