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Find the standard form of the equation of the circle with center (-3, 1) and tangent to the y-axis.
Let me see.
If the circle is tangent to the y-axis, then it must touch the line x = 0 at the point (0, 1).
I need to find r.
r = sqrt{(0 - (-3))^2 + (1 - 1)^2}
r = 3
x^2 + y^2 = r^2
x^2 + y^2 = (3)^2
The equation is x^2 + y^2 = 9.
You say?
Last edited by nycguitarguy (2024-03-02 18:08:12)
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You did well for r by applying the right equation of a circle. Then... I didn't see it.
I guess you know that:
x^2 + y^2 = r^2
means that the circle center is at (0, 0).
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You did well for r by applying the right equation of a circle. Then... I didn't see it.
I guess you know that:
x^2 + y^2 = r^2
means that the circle center is at (0, 0).
Is my answer right?
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KerimF wrote:You did well for r by applying the right equation of a circle. Then... I didn't see it.
I guess you know that:
x^2 + y^2 = r^2
means that the circle center is at (0, 0).Is my answer right?
But... your exercise is to find the standard form of the equation of the circle with center (-3, 1) ... not (0,0)
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FelizNYC wrote:KerimF wrote:You did well for r by applying the right equation of a circle. Then... I didn't see it.
I guess you know that:
x^2 + y^2 = r^2
means that the circle center is at (0, 0).Is my answer right?
But... your exercise is to find the standard form of the equation of the circle with center (-3, 1) ... not (0,0)
Ok. Copy.
I found r to be 3.
(x - h)^2 + (y - k)^2 = r^2
(x - (-3))^2 + (y - 1)^2 = 3^2
(x + 3)^2 + (y - 1)^2 = 9
You say?
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It is right.
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It is right.
Beautiful. Cool. It feels good to be right.
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