AP EAPCET 04-Jul-2022 Shift 2 Paper

(d) To find the shortest and longest distances from the point $(2,-7)$ to the circle $x^2+y^2-14 x-10 y-151=0$So, 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 $O M=O N=$ radius Thus, the shortest distance $P M \dot{M}=O M-O P$ and the longest distance $P N=M N-P M$First 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 $x^2+y^2-14 x-10 y-151$$=(2)^2+(-7)^2-14 \times 2-10(-7)-151=-56< 0$Thus, the point lies inside of the circle. Thus, we know any equation of the circle is in the from $x^2+y^2+2 g x+2 f y+c=0$Thus, $g=-7$, and $f=-5, c=-151$So, centre is $(-g,-f)=(7,5)$and radius $=r=O M=O N$.$\;=\sqrt{7^2+5^2-(-151)}$$\;=\sqrt{49+25+151}$$\;=\sqrt{225}=15$Now, distance between centre $O(7,5)$ and $P(2,-7)$ is $O P=\sqrt{(7-2)^2+(5+7)^2}$$=\sqrt{25+144}=\sqrt{169}=13$Thus, the shortest distance from the point $P(2,-7)$ to the circle is $P M=O M-O P=15-13=2$ units and the longest distance is$P N=M N-P M\;=(O M+O N)-P M$$\;=(15+15)-2=28 \;\text{ units }\;$Thus, the required ratio is $28 ; 2=14: 1$