Question

Write the equation in standard form for a circle with center (-8,9) and tangent to the x-axis.

Answer 1

Answer:

The standard form of the equation of the circle is $(x+8)^{2}+(y-7)^{2}=49$

Solution:

Given, a circle has center (-8, 9) and x – axis as tangent.

$\text { The standard form is } x^{2}+y^{2}=r^{2}$

First, let's determine r.  

The center is 9 above the x-axis and the circle is tangent to the x-axis, so the radius r should be equal to 9.

The center moved 8 to the left, so substitute x by (x + 8).

The center moved 9 up, so substitute y by (y − 9).

You can determine these numbers by filling in the center coordinates, the outcome must be zero. [x+8=−8+8=0]

The standard equation $(x-a)^{2}+(y-b)^{2}=r^{2} \text { for a circle with center }(a, b)$

So, the equation becomes

$(x+8)^{2}+(y-7)^{2}=7^{2} \rightarrow(x+8)^{2}+(y-7)^{2}=49$

Hence, standard form of the circle is $(x+8)^{2}+(y-7)^{2}=49$