The circle touching the coordinate axes with | vedclass

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

Question

(A) Since the circle touches both coordinate axes,its centre $(h, k)$ must satisfy $|h| = |k| = r$,where $r$ is the radius.Thus,the centre is either $

Vedclass · Accepted Answer

$x^2+y^2-2x+2y+1=0$

Vedclass · Answer

(A) Since the circle touches both coordinate axes,its centre $(h, k)$ must satisfy $|h| = |k| = r$,where $r$ is the radius.Thus,the centre is either $(r, r), (r, -r), (-r, r),$ or $(-r, -r)$.The centre lies on the line $x-2y-3=0$.Case $1$: Centre is $(r, r)$. Then $r-2r-3=0$ $\Rightarrow -r=3$ $\Rightarrow r=-3$. Since $r > 0$,this is not possible.Case $2$: Centre is $(r, -r)$. Then $r-2(-r)-3=0$ $\Rightarrow 3r=3$ $\Rightarrow r=1$. The centre is $(1, -1)$.The equation of the circle is $(x-1)^2+(y+1)^2=1^2$ $\Rightarrow x^2-2x+1+y^2+2y+1=1$ $\Rightarrow x^2+y^2-2x+2y+1=0$.Case $3$: Centre is $(-r, r)$. Then $-r-2r-3=0$ $\Rightarrow -3r=3$ $\Rightarrow r=-1$. Not possible.Case $4$: Centre is $(-r, -r)$. Then $-r-2(-r)-3=0 \Rightarrow r=3$. The centre is $(-3, -3)$.The equation of the circle is $(x+3)^2+(y+3)^2=3^2$ $\Rightarrow x^2+6x+9+y^2+6y+9=9$ $\Rightarrow x^2+y^2+6x+6y+9=0$ (not in options).Therefore,the correct equation is $x^2+y^2-2x+2y+1=0$.

The circle touching the coordinate axes with its centre lying on $x-2y-3=0$ is

Similar Questions

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:

The line $4x + 3y - 4 = 0$ divides the circumference of a circle in the ratio $1:2$. If $C(5, 3)$ is the center of that circle,then the equation of the circle is

If $(6, -k)$ and $(-3, 2)$ are conjugate points with respect to the circle $x^2 + y^2 + 6x + 4y + 12 = 0$,then $k$ equals:

Consider the circle $x^2+y^2-4x-2y+c=0$ whose centre is $A(2,1)$. If the point $P(10,7)$ is such that the line segment $PA$ meets the circle in $Q$ with $PQ=5$,then $c$ is equal to

Let $C$ be a circle with radius $\sqrt{10}$ units and centre at the origin. Let the line $x+y=2$ intersect the circle $C$ at the points $P$ and $Q$. Let $MN$ be a chord of $C$ of length $2$ units and slope $-1$. Then,the distance (in units) between the chord $PQ$ and the chord $MN$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module