If lambda is the perpendicular distance of a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given equation of the circle is $x^2+y^2+2x+2y-3=0$. Completing the square,we get $(x+1)^2+(y+1)^2=5$. This is a circle with center $C(-1, -1)

Vedclass · Accepted Answer

$3 \sqrt{5}$

Vedclass · Answer

(B) The given equation of the circle is $x^2+y^2+2x+2y-3=0$. Completing the square,we get $(x+1)^2+(y+1)^2=5$. This is a circle with center $C(-1, -1)$ and radius $r = \sqrt{5}$. The distance $d$ from the center $C(-1, -1)$ to the line $2x+y+13=0$ is given by $d = \frac{|2(-1) + (-1) + 13|}{\sqrt{2^2 + 1^2}} = \frac{|-2-1+13|}{\sqrt{5}} = \frac{10}{\sqrt{5}} = 2\sqrt{5}$. The maximum distance of a point on the circle from the line is $d + r$. Thus,$\lambda_{max} = 2\sqrt{5} + \sqrt{5} = 3\sqrt{5}$.

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

Similar Questions

Find the equation of the director circle of the circle $x^2 + y^2 = 8$.

The radius of the circle having $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$ as its tangents is

If the points $P(3, 1)$ and $Q(6, 5)$ form a triangle with a third point $R(x, y)$ such that the area of $\Delta PQR$ is $6$ square units and $\angle PRQ = \frac{\pi}{2}$,then the number of possible positions for point $R$ is:

Among the chords of the circle $x^2+y^2=75$,the number of chords having their midpoints on the line $x=8$ and having their slopes as integers is

The chord of the circle $x^{2}+y^{2}-4x=0$ which is bisected at $(1,0)$ is perpendicular to the line

Vedclass Test Series

Exam Paper Generator

Online Exam Module