A circle has its center in Quadrant I. It is tangent to the y-axis at (0,2) and its radius is 3. What is the equation of the cir
cle? Explain the process you use to find the equation.
1 answer:
We know that
the equation of a circle is
(x-h)²+(y-k)²=r²
r=3 units
where the center is the point (h,k)
if the circle is tangent to the y-axis at (0,2) and <span> its center in Quadrant I
</span><span>then
the center is the point (0+3,2)--------> (3,2)
therefore
the answer is
</span>(x-3)²+(y-2)²=9
see the attached figure
the equation of a circle is
(x-h)²+(y-k)²=r²
r=3 units
where the center is the point (h,k)
if the circle is tangent to the y-axis at (0,2) and <span> its center in Quadrant I
</span><span>then
the center is the point (0+3,2)--------> (3,2)
therefore
the answer is
</span>(x-3)²+(y-2)²=9
see the attached figure
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