Q.46 If \lambda is the perpendicular distance of a point P on the circle x^{2}+y^{2}+2 x+2 y-3=0, from the line 2 x+y+13=0, then maximum possible value of \lambda is
Options:
A. 2 \sqrt{5}
B. 3 \sqrt{5}
C. 4 \sqrt{5}
D. \sqrt{5}
Answer: B
Solution:
Given equation of the circle is
x^{2}+y^{2}+2 x+2 y-3=0Which can be written as: (x+1)^{2}+(y+1)^{2}=5
It is a circle with centre (-1,-1) and radius \sqrt{5}
Given line is: 2 x+y+13=0
To find the required distance, we find the equation of a line perpendicular to the given line, and passing through the centre of the given circle.
\therefore \quad Equation of this line is: (y+1)=\frac{1}{2}(x+1) i.e., x=2 y+1
Now, we find the points where line x=2 y+1 intersects the circle x^{2}+y^{2}+2 x+2 y-3=0
\therefore(2 y+1)^{2}+y^{2}+2(2 y+1)+2 y-3=0 \therefore 4 y^{2}+4 y+1+y^{2}+4 y+2+2 y-3=0 \therefore 5 y^{2}+10 y=0 \therefore \quad y(y+2)=0\therefore \quad y=0 or y=-2
\therefore \quad x=1 or x=-3
\therefore \quad(1,0) and (-3,-2) are the points on the circle, and one of them is at the maximum distance from the given line.
\therefore \quad \mathrm{d}_{1}=\left|\frac{2(1)+(0)+13}{\sqrt{4+1}}\right| and \mathrm{d}_{2}=\left|\frac{2(-3)+(-2)+13}{\sqrt{4+1}}\right|
\therefore \quad \mathrm{d}_{1}=\frac{15}{\sqrt{5}}=3 \sqrt{5} \quad and \mathrm{d}_{2}=\frac{5}{\sqrt{5}}=\sqrt{5}
\therefore \lambda=3 \sqrt{5}