What is the standard form of the equation of a circle with center (-3,3) and tangent to the line y=1?

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What is the standard form of the equation of a circle with center (-3,3) and tangent to the line y=1?

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Answer 1 · 2016-05-07T13:23:39

Equation of circle is $x^{2} + y^{2} + 6 x - 6 y + 14 = 0$ $x^2+y^2+6x-6y+14=0$ and $y = 1$ $y=1$ is tangent at $(- 3, 1)$ $(-3,1)$

Explanation:

The equation of a circle with center $(- 3, 3)$ $(-3,3)$ with radius $r$ $r$ is

${(x + 3)}^{2} + {(y - 3)}^{2} = r^{2}$ $(x+3)^2+(y-3)^2=r^2$

or $x^{2} + y^{2} + 6 x - 6 y + 9 + 9 - r^{2} = 0$ $x^2+y^2+6x-6y+9+9-r^2=0$

As $y = 1$ $y=1$ is a tangent to this circle, putting $y = 1$ $y=1$ in the equation of a circle should give only one solution for $x$ $x$ . Doing so we get

$x^{2} + 1 + 6 x - 6 + 9 + 9 - r^{2} = 0$ $x^2+1+6x-6+9+9-r^2=0$ or

$x^{2} + 6 x + 13 - r^{2} = 0$ $x^2+6x+13-r^2=0$

and as we should have only one solution, discriminant of this quadratic equation should be $0$ $0$ .

Hence, $6^{2} - 4 \times 1 \times (13 - r^{2}) = 0$ $6^2-4xx1xx(13-r^2)=0$ or

$36 - 52 + 4 r^{2} = 0$ $36-52+4r^2=0$ or $4 r^{2} = 16$ $4r^2=16$ and as $r$ $r$ has to be positive

$r = 2$ $r=2$ and hence equation of circle is

$x^{2} + y^{2} + 6 x - 6 y + 9 + 9 - 4 = 0$ $x^2+y^2+6x-6y+9+9-4=0$ or $x^{2} + y^{2} + 6 x - 6 y + 14 = 0$ $x^2+y^2+6x-6y+14=0$

and $y = 1$ $y=1$ is tangent at $(- 3, 1)$ $(-3,1)$

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What is the standard form of the equation of a circle with center (-3,3) and tangent to the line y=1?

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