Two circles which touch both the coordinate a | vedclass

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

Question

(D) Let the radius of a circle touching both coordinate axes be $r$. The equation of such a circle is $(x-r)^2 + (y-r)^2 = r^2$. Since the circle pass

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(D) Let the radius of a circle touching both coordinate axes be $r$. The equation of such a circle is $(x-r)^2 + (y-r)^2 = r^2$. Since the circle passes through $A=(1,2)$,we have $(1-r)^2 + (2-r)^2 = r^2$. Expanding this,we get $1 - 2r + r^2 + 4 - 4r + r^2 = r^2$,which simplifies to $r^2 - 6r + 5 = 0$. Solving for $r$,we get $(r-1)(r-5) = 0$,so $r=1$ or $r=5$. The two circles are $C_1: (x-1)^2 + (y-1)^2 = 1$ and $C_2: (x-5)^2 + (y-5)^2 = 25$. The common chord $AB$ is given by $C_1 - C_2 = 0$. $C_1: x^2 + y^2 - 2x - 2y + 1 = 0$ $C_2: x^2 + y^2 - 10x - 10y + 25 = 0$ Subtracting $C_2$ from $C_1$: $(x^2 - x^2) + (y^2 - y^2) + (-2x + 10x) + (-2y + 10y) + (1 - 25) = 0$,which gives $8x + 8y - 24 = 0$,or $x + y = 3$. Since $A=(1,2)$ lies on $x+y=3$,the point $B$ is the reflection of $A$ across the line connecting the centers $(1,1)$ and $(5,5)$. The line of centers is $y=x$. The reflection of $(1,2)$ across $y=x$ is $(2,1)$. Thus $B=(2,1)$. The length $AB = \sqrt{(2-1)^2 + (1-2)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.

Two circles which touch both the coordinate axes intersect at the points $A$ and $B$. If $A=(1,2)$,then $AB=$

Similar Questions

The equation of the circle whose diameter is the common chord of the circles $x^2+y^2+2x+3y+1=0$ and $x^2+y^2+4x+3y+2=0$ is

The equation of the common chord of the circles $x^2+y^2+2x+3y+1=0$ and $x^2+y^2-5x-6y+4=0$ is

$y = mx$ is a chord of a circle of radius $a$. The diameter of the circle lies along the $x$-axis and one end of this chord is at the origin. The equation of the circle described on this chord as diameter is

From a point $P(-4, 0)$,two tangents are drawn to the circle $x^2 + y^2 - 4x - 6y - 12 = 0$ touching the circle at $A$ and $B$. If the equation of the circle passing through $P, A$,and $B$ is $x^2 + y^2 + 2gx + 2fy + c = 0$,then $(g, f) =$

If the radical axis of the circles $x^2+y^2+2gx+2fy+c=0$ and $2x^2+2y^2+3x+8y+2c=0$ touches the circle $x^2+y^2+2x+2y+1=0$,then $(4g-3)(f-2)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module