(A) Since the circle touches both axes and lies in the fourth quadrant with radius $r = 1$,the center of the circle must be $(1, -1)$.The standard equ

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(A) Since the circle touches both axes and lies in the fourth quadrant with radius $r = 1$,the center of the circle must be $(1, -1)$.The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.Substituting $h = 1$,$k = -1$,and $r = 1$:$(x - 1)^2 + (y - (-1))^2 = 1^2$$(x - 1)^2 + (y + 1)^2 = 1$Expanding the equation:$(x^2 - 2x + 1) + (y^2 + 2y + 1) = 1$$x^2 + y^2 - 2x + 2y + 2 = 1$$x^2 + y^2 - 2x + 2y + 1 = 0$

Class 11 Mathematics 10-1.Circle and System of Circles Equations of circle

Easy

$A$ circle touches both axes and its center lies in the fourth quadrant. If its radius is $1$,then its equation is:

A
$x^2 + y^2 - 2x + 2y + 1 = 0$
B
$x^2 + y^2 + 2x - 2y + 1 = 0$
C
$x^2 + y^2 - 2x - 2y + 1 = 0$
D
$x^2 + y^2 + 2x + 2y + 1 = 0$

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10-1.Circle and System of Circles

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Equations of circle

If the lines $x + y = 6$ and $x + 2y = 4$ are diameters of a circle whose diameter is $20$,then the equation of the circle is:

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The straight line $2x + 3y - k = 0, k > 0$ cuts the $x$ and $y$-axes at $A$ and $B$ respectively. The area of $\triangle OAB$,where $O$ is the origin,is $12 \text{ sq unit}$. The equation of the circle having $AB$ as diameter is

DifficultKCET 2010

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If the equation $px^2 + (2 - q)xy + 3y^2 - 6qx + 30y + 6q = 0$ represents a circle,then the values of $p$ and $q$ are:

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Let the circle $S = x^2 + y^2 + 2gx + 2fy + c = 0$ touch the positive $X$-axis and the positive $Y$-axis. Let $(2, 4)$ be a point on the circle $S = 0$. If two such circles exist, then the difference of their areas is (in $\pi$)

MediumTS EAMCET 2023

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$A$ circle passes through the point $(1, -2)$ and touches the $x$-axis at $(3, 0)$. Through which of the following points does it also pass?

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$A$ circle touches both axes and its center lies in the fourth quadrant. If its radius is $1$,then its equation is:

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If the lines $x + y = 6$ and $x + 2y = 4$ are diameters of a circle whose diameter is $20$,then the equation of the circle is:

The straight line $2x + 3y - k = 0, k > 0$ cuts the $x$ and $y$-axes at $A$ and $B$ respectively. The area of $\triangle OAB$,where $O$ is the origin,is $12 \text{ sq unit}$. The equation of the circle having $AB$ as diameter is

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