(d) To find the shortest and longest distances from the point
(2,−7) to the circle
x2+y2−14x−10y−151=0So, to find the solution we first draw the diagram,
So, we have to find the coordinates of the centre 0 and the radius of the circle, i.e
OM=ON= radius Thus, the shortest distance
PM˙M=OM−OP and the longest distance
PN=MN−PMFirst we have to determine the point
P(2,−7) lies in which side of circle. So, we put the value
x=2,
y=−7 in the left hand side of the equation, we get
x2+y2−14x−10y−151=(2)2+(−7)2−14×2−10(−7)−151=−56<0Thus, the point lies inside of the circle. Thus, we know any equation of the circle is in the from
x2+y2+2gx+2fy+c=0Thus,
g=−7, and
f=−5,c=−151So, centre is
(−g,−f)=(7,5)and radius
=r=OM=ON.
‌=√72+52−(−151)‌=√49+25+151‌=√225=15Now, distance between centre
O(7,5) and
P(2,−7) is
OP=√(7−2)2+(5+7)2=√25+144=√169=13Thus, the shortest distance from the point
P(2,−7) to the circle is
PM=OM−OP=15−13=2 units and the longest distance is
PN=MN−PM‌=(OM+ON)−PM‌=(15+15)−2=28‌ units ‌Thus, the required ratio is
28;2=14:1