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mathdad
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Find the equation of the circle tangent to the y-axis and with center (3, 5).
Can someone provide the steps needed to solve this problem?
Can someone provide the steps needed to solve this problem?
MarkFL said:The equation of a circle centered ar $(h,k)$ is given by:
\(\displaystyle (x-h)^2+(y-k)^2=r^2\)
If the circle is tangent to the $y$-axis, then its radius must be $r=|h|\implies r^2=h^2$, thus we have:
\(\displaystyle (x-h)^2+(y-k)^2=h^2\)
We are given $(h,k)=(3,5)$, so plug in those numbers. :D
The standard form of the equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h,k) is the center of the circle and r is the radius.
To find the center and radius of a circle given its equation, first rewrite the equation in standard form. Then, the values of h and k will give the coordinates of the center, and the square root of r^2 will give the radius.
Yes, the equation of a circle can have fractions or decimals. When graphing, these values will result in a non-integer value for the radius, but the equation will still represent a circle.
To determine if a point (x,y) lies inside, outside, or on the circle, plug in the values of x and y into the equation of the circle. If the resulting value is equal to r^2, the point lies on the circle. If it is greater than r^2, the point lies outside the circle. If it is less than r^2, the point lies inside the circle.
Yes, the equation of a circle can have negative values for h or k. This simply means that the center of the circle lies in a quadrant other than the first quadrant. The absolute values of h and k will still give the coordinates of the center.