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
You might be interested in
Answer:
The answer is "3".
Step-by-step explanation:
If the compression factor of 3 is modified, the k value is 3 For a function f(x) the horizontal extending or compression is provided by g = f (bx)
Here b is constant. Where b is constant.
If b> 0, the function graph is condensed. The fact that perhaps the function is transformed by an encoding factor of 3 is given in the question.