If the straight line 4x + 3y + lambda = 0 tou | vedclass

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

Question

(D) The given equation of the circle is $2(x^2 + y^2) = 5$,which can be rewritten as $x^2 + y^2 = \frac{5}{2}$.Comparing this with the standard form $

Vedclass · Accepted Answer

$\frac{5\sqrt{10}}{2}$

Vedclass · Answer

(D) The given equation of the circle is $2(x^2 + y^2) = 5$,which can be rewritten as $x^2 + y^2 = \frac{5}{2}$.Comparing this with the standard form $x^2 + y^2 = r^2$,the radius $r = \sqrt{\frac{5}{2}}$.The center of the circle is $(0, 0)$.$A$ line $ax + by + c = 0$ touches a circle with center $(h, k)$ and radius $r$ if the perpendicular distance from the center to the line equals the radius.Thus,$\frac{|4(0) + 3(0) + \lambda|}{\sqrt{4^2 + 3^2}} = \sqrt{\frac{5}{2}}$.$\frac{|\lambda|}{\sqrt{16 + 9}} = \sqrt{\frac{5}{2}}$.$\frac{|\lambda|}{5} = \sqrt{\frac{5}{2}}$.$|\lambda| = 5 	imes \frac{\sqrt{5}}{\sqrt{2}} = \frac{5\sqrt{5}}{\sqrt{2}} 	imes \frac{\sqrt{2}}{\sqrt{2}} = \frac{5\sqrt{10}}{2}$.Therefore,$\lambda = \pm \frac{5\sqrt{10}}{2}$.

If the straight line $4x + 3y + \lambda = 0$ touches the circle $2(x^2 + y^2) = 5$,then $\lambda$ is

Similar Questions

Which of the following lines is a tangent to the circle $x^2 + y^2 = 25$ for all values of $m$?

The equation of the line passing through the point $(x', y')$ and perpendicular to the line $yy' = 2a(x + x')$ is

The equation of the common tangent touching the circle $(x-3)^2+y^2=9$ and the parabola $y^2=4x$ above the $X$-axis is

Abscissae of points on the curve $xy = (c + x)^2$,the normal at which cuts off numerically equal intercepts from the axes of coordinates is/are:

Let $A$ be the centre of the circle $x^{2}+y^{2}-2x-4y-20=0$. Let $B(1,7)$ and $D(4,-2)$ be two points on the circle such that tangents at $B$ and $D$ meet at $C$. The area of the quadrilateral $ABCD$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module