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3 years ago

Which equation describes the circle having center point (3,7) and the radius r=4 in Standard form

Mathematics

1 answer:

Annette [7]3 years ago

6 0

The equation of circle in standard form is $(x - 3)^2 + (y - 7)^2 = 16$

<h3><u>Solution:</u></h3>

Given that circle having center point (3,7) and the radius r = 4

To find: equation of circle in standard form

<em><u>The equation of circle is given as:</u></em>

$(x - h)^2 + (y - k)^2 = r^2$

Where center (h,k) and radius r units

Given that center point (h , k) = (3, 7) and radius r = 4 units

Substituting the values in above equation of circle,

$(x - 3)^2 + (y - 7)^2 = 4^2\\\\(x - 3)^2 + (y - 7)^2 = 16$

Thus the equation of circle in standard form is $(x - 3)^2 + (y - 7)^2 = 16$

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3) if its vertex is above the x-axis, it will have NO real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently.

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Since in our case $a=2$ and $b=-3$ , we get that the x-position of the vertex is:

$x_v=\frac{-b}{2a}\\x_v=\frac{-(-3)}{2(2)}\\x_v=\frac{3}{4}$

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$y_v=f(\frac{3}{4})=2( \frac{3}{4})^2-3(\frac{3}{4})+4\\f(\frac{3}{4})=2( \frac{9}{16})-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{18}{8}+\frac{32}{8}\\f(\frac{3}{4})=\frac{23}{8}$

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See the graph produced in the attached image.

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