The area of the circle having its centre at ( | vedclass

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

Question

(A) Given,the centre of the circle is $C(3,4)$.Since the circle touches the line $5x+12y-11=0$,the radius $r$ of the circle is equal to the perpendicu

Vedclass · Accepted Answer

$16 \pi$ sq units

Vedclass · Answer

(A) Given,the centre of the circle is $C(3,4)$.Since the circle touches the line $5x+12y-11=0$,the radius $r$ of the circle is equal to the perpendicular distance from the centre $(3,4)$ to the line.The perpendicular distance $d$ from a point $(x_1, y_1)$ to the line $ax+by+c=0$ is given by $d = \frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$.Substituting the values,we get:$r = \frac{|5(3)+12(4)-11|}{\sqrt{5^2+12^2}}$$r = \frac{|15+48-11|}{\sqrt{25+144}}$$r = \frac{|52|}{\sqrt{169}}$$r = \frac{52}{13} = 4$.The area of the circle is given by $\pi r^2$.Area $= \pi(4)^2 = 16\pi$ sq units.

The area of the circle having its centre at $(3,4)$ and touching the line $5x+12y-11=0$ is

Similar Questions

The least and the greatest distances of the point $(10, 7)$ from the circle $x^{2} + y^{2} - 4x - 2y - 20 = 0$ are

The length of the diameter of a circle which touches the $x$-axis at the point $(1, 0)$ and passes through the point $(2, -3)$ is:

$P(3,1)$,$Q(6,5)$ and $R(x,y)$ are three points such that the angle $\angle PRQ$ is a right angle and the area of the $\Delta RPQ = 5$. Then the number of such points $R$ is

Let the normals at all the points on a given curve pass through a fixed point $(a, b)$. If the curve passes through $(3, -3)$ and $(4, -2\sqrt{2})$,and given that $a - 2\sqrt{2}b = 3$,then $(a^{2} + b^{2} + ab)$ is equal to ..... .

$A$ circle with center at $(2,4)$ is such that the line $x+y+2=0$ cuts a chord of length $6$. The radius of the circle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module