Circles
5.0 Maximum and minimum distance of a point from a circle
5.0 Maximum and minimum distance of a point from a circle
Let us assume a point $P({x_1},{y_1})$ and equation of circle be $S \equiv {x^2} + {y^2} + 2gx + 2fy + c = 0$ with centre $C(-g,-f)$ and radius $r = \sqrt {{g^2} + {f^2} - c} $.
CASE I: If point $P$ lies inside the circle i.e., ${S_1} < 0$ and radius $CA = \sqrt {{g^2} + {f^2} - c} = CB$ as shown in figure $14$.
Minimum distance of $P$ from circle$ = PA = CP - CA = \left| {CP - } \right|$.
Maximum distance of $P$ from circle$ = PB = CB + CP = \left| {r + CP} \right|$.
CASE II: If point $P$ lies outside the circle i.e., ${S_1} > 0$ and radius $CA = \sqrt {({g^2} + {f^2} - c)} = CB$ as shown in figure $15$.
Minimum distance of $P$ from circle$ = PA = CP - CA = \left| {CP - r} \right|$.
Maximum distance of $P$ from circle$ = PB = CP + CB = \left| {CP + r} \right|$.
CASE III: If point $P$ lies on the circle i.e., ${S_1} = 0$ and radius $CA = \sqrt {({g^2} + {f^2} - c)} = CB$ as shown in figure $16$.
Minimum distance of $P$ from circle$=0$.
Maximum distance of $P$ from circle$ = AB = PB = 2r$.
Question 11. If the point $P(\lambda , - \lambda )$ lies inside the circle ${x^2} + {y^2} - 4x + 2y - 8 = 0$, then find the range of $\lambda $.
Solution: If point $P({x_1},{y_1})$ lies inside the circle, then ${x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c < 0$
From the values given in the question, ${x_1} = \lambda $ and ${y_1} = - \lambda $
Therefore, ${\text{}}{\lambda ^2} + {\left( { - {\text{}}\lambda } \right)^2} - 4\lambda - 2\lambda - 8 < 0$
or, $${\text{}}2{\lambda ^2} - 6\lambda - 8 < 0$$ $${\text{}}{\lambda ^2} - 3\lambda - 4 < 0$$ $${\text{}}{\lambda ^2} - 4\lambda + \lambda - 4 < 0$$ $$\lambda \left( {\lambda - 4} \right) + 1\left( {\lambda - 4} \right) < 0$$ $$\left( {\lambda - 4} \right)\left( {\lambda + 1} \right) < 0$$ $$\lambda < 4\ or\ \lambda > - 1$$
Therefore, range of $\lambda $ is $(-1,4)$.
Question 12. A point $P(2,-7)$ lies inside the circle ${x^2} + {y^2} - 14x - 10y - 151 = 0$. Find the maximum and minimum distance of $P$ from the circle and also find the coordinates of $A$ and $B$ as shown in figure $E1$.
Solution: From the equation of circle given, $g = - 7,{\text{ }}f = - 5,{\text{ }}c = - 151$
Radius of circle $r = \sqrt {{g^2} + {f^2} - c} = \sqrt { - {7^2} + - {5^2} + 151} = \sqrt {49 + 21 + 151} = \sqrt {225} = 15$
Centre of circle is $(-g,-f)$ i.e., $(7,5)$ and coordinates of point $P(x{\text{'}},y{\text{'}})$ is $(2,-7)$
Distance between centre $C$ and point $P$ using distance formulae is $$CP = \sqrt {{{\left( {x' + g} \right)}^2} + {{\left( {y' + f} \right)}^2}} = \sqrt {{{\left( {2 - 7} \right)}^2} + {{\left( { - 7 - 5} \right)}^2}} = \sqrt {{{\left( { - 5} \right)}^2} + {{\left( { - 12} \right)}^2}} $$$$ = \sqrt {25 + 144} = \sqrt {169} = 13$$
Maximum distance of $P$ from the circle$ = PB = CP + r = 13 + 15 = 28$
Minimum distance of$P$ from the circle$ = AP = r - CP = 15 - 13 = 2$
Apply section formulae between points $A$ and $C$ in figure $E2$, we get 
$2 = \frac{{13{x_1} + 14}}{{15}}$ and $ - 7 = \frac{{13{y_1} + 10}}{{15}}$
$30 - 14 = 13{x_1}$ and $ - 105 - 10 = 13{y_1}$
Therefore, ${x_1} = \frac{{16}}{{13}}$ and ${y_1} = \frac{{ - 115}}{{13}}$
Apply section formulae between points $P$ and $B$ in figure $E2$, we get
$7 = \frac{{13{x_2} + 30}}{{28}}$ and $5 = \frac{{13{y_2} - 105}}{{28}}$
$196 - 30 = 13{x_2}$ and $140 + 105 = 13{y_2}$
Therefore, ${x_2} = \frac{{166}}{{13}}$ and ${y_2} = \frac{{245}}{{13}}$.
