Question

Let
( S=x^{2}+y^{2}-14 x-10 y-151=0 )
Substituting the point (2,-7) in this
equation, we get
( S_{1}=(2)^{2}+(-7)^{2}-14(2)-10(-7) )
( -151=-56<0 )
( therefore P(2,-7) ) inside the circle
Radius of circle
( r=sqrt{(-7)^{2}+(-5)^{2}+151}=15 )
Centre of circle (7,5)
( mathrm{CP}=sqrt{(7-2)+(5+7)^{2}}=13 )
Shortest distance ( = )
( P A=r-C P=15-13=2 )
Largest distance ( P B=r+C P=15+13=28 )

# PUPIENUNCI Find the shortest and largest distance from the point (2, - 7) to the circle x2 + y2 - 14x - 10y - 151 = 0

Solution