If the equation of the circle lying in the fi | vedclass

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

Question

(C) Since the circle lies in the first quadrant and touches both coordinate axes,its center is $(c, c)$ and its radius is $c$,where $c > 0$.The equati

Vedclass · Accepted Answer

$1$ or $6$

Vedclass · Answer

(C) Since the circle lies in the first quadrant and touches both coordinate axes,its center is $(c, c)$ and its radius is $c$,where $c > 0$.The equation of the circle is $(x-c)^2 + (y-c)^2 = c^2$.The line is $\frac{x}{3} + \frac{y}{4} = 1$,which can be written as $4x + 3y - 12 = 0$.Since the circle touches this line,the perpendicular distance from the center $(c, c)$ to the line must be equal to the radius $c$.$\frac{|4c + 3c - 12|}{\sqrt{4^2 + 3^2}} = c$$\frac{|7c - 12|}{5} = c$$|7c - 12| = 5c$Case $1$: $7c - 12 = 5c \implies 2c = 12 \implies c = 6$.Case $2$: $7c - 12 = -5c \implies 12c = 12 \implies c = 1$.Thus,$c = 1$ or $c = 6$.

If the equation of the circle lying in the first quadrant,touching both the coordinate axes and the line $\frac{x}{3}+\frac{y}{4}=1$ is $(x-c)^2+(y-c)^2=c^2$,then $c=$

Similar Questions

If $(4,7)$ and $(-2,-1)$ are ends of a diameter of a circle which intersects $X$-axis at $A$ and $B$,then $AB$ is equal to

If $(3,1)$ and $(-2,4)$ are points on a circle $S$ whose centre lies on the line $x-y+1=0$,then the parametric equations of $S$ are:

The equation of the circle touching the lines $|x-2|+|y-3|=4$ is

The equations of the circle which pass through the origin and make intercepts of lengths $4$ and $8$ on the $x$ and $y$-axes respectively are

If a circle passing through $(4, -2)$ is concentric with the circle $x^2 + y^2 - 2x + 4y + 20 = 0$,then find the value of $c$ for the circle $x^2 + y^2 - 2x + 4y + c = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module