A) 20
B) 12
C) 10
D) 16
E) 22
Correct Answer: A
Solution :
Given equation of circle \[{{x}^{2}}+{{y}^{2}}+4x-10y-7=0\] with centre\[=(-2,5)=C\]and radius\[=\sqrt{4+25+7}=\sqrt{36}=6\] Now, \[PC=\sqrt{{{(4+2)}^{2}}+{{(-3-5)}^{2}}}\] \[=\sqrt{36+64}=\sqrt{100}=10\] So, the point lies outside the circle. Now, maximum distance \[=PQ=PC+QC=10+6=16\] and minimum distance \[=PS=PC-SC=10-6=4\] \[\therefore \]Sum of maximum and minimum distance \[=16+4=20\]You need to login to perform this action.
You will be redirected in
3 sec