Q.46. If \lambda is the perpendicular distance of a point P on the circle x^2 + y^2 + 2x + 2y - 3 = 0 , from the line 2x + y + 13 = 0 , then the maximum possible value of \lambda is
A. 2\sqrt{5}
B. 3\sqrt{5}
C. 4\sqrt{5}
D. \sqrt{5}
Answer :- B. 3\sqrt{5}
Explanation :- Given equation of the circle is
x2+y2+2x+2y-3= 0
which can be writtten as (x+1)2 +(y+1)2 = 5
It is a circle with center (-1, -1) and radius \sqrt{5}
Given line is: 2x + y + 13 = 0
To find the required distance, we find the equation of a line perpendicular to the given line, passing through the center of the given circle.
Equation of this line is: (y + 1) = \frac{1}{2}(x + 1) , i.e., x = 2y + 1
Now, we find the points where the line x = 2y + 1 intersects the circle x^2 + y^2 + 2x + 2y - 3 = 0
(2y + 1)^2 + y^2 + 2(2y + 1) + 2y - 3 = 0 4y^2 + 4y + 1 + y^2 + 4y + 2 + 2y - 3 = 0 5y^2 + 10y = 0 y(y + 2) = 0 y = 0 \text{ or } y = -2 x = 1 \text{ or } x = -3Points on the circle: (1, 0) and (-3, -2) , and one of them is at the maximum distance from the line.
d_1 = \left| \frac{2(1) + 0 + 13}{\sqrt{4 + 1}} \right| and d_2 = \left| \frac{2(-3) + (-2) + 13}{\sqrt{4 + 1}} \right|
d_1 = \frac{15}{\sqrt{5}} = 3\sqrt{5} and d_2 = \frac{5}{\sqrt{5}} = \sqrt{5}
\lambda = 3\sqrt{5}