<p>1) Find the center and radius of the circle whose equation is x^2+7x+y^2-10y-
ID: 3099863 • Letter: #
Question
<p>1) Find the center and radius of the circle whose equation is x^2+7x+y^2-10y-15=0<br /><br />center of the circle is (___, ____)<br />the radius of the circle is ____<br /><br />the same quation in the form (x-h)^2+(y-k)^2=r^2<br /><br />____+ (y-5)^2 =______</p><p> </p>
<p>2) Find the center and radius of the circle whose equation is <img src="https://webwork.uwstout.edu/webwork2_files/tmp/equations/1f/5e6c37cc8e4701dc98ef6b05ab0b3a1.png" alt="x^2 + 7 x + y^2 - 10 y - 15 = 0" />. <br />The center of the circle is (<input name="AnSwEr1" size="5" type="TEXT" value="49/4" /> , <input name="AnSwEr2" size="5" type="TEXT" value="-5" />) <br />The same equation in the form <img src="https://webwork.uwstout.edu/webwork2_files/tmp/equations/dc/debb9fef17c9413af2152cdfa953e61.png" alt="(x-h)^2 + (y-k)^2 = r^2" /> is <br /><input name="AnSwEr4" size="10" type="TEXT" value="(x+49/4)^2" /> <img src="https://webwork.uwstout.edu/webwork2_files/tmp/equations/04/38b61e1c4cbccf395cd43978aeb0751.png" alt="+" /> <input name="AnSwEr5" size="10" type="TEXT" value="(y-5)^2" /> <img src="https://webwork.uwstout.edu/webwork2_files/tmp/equations/cf/53c4dbf80d3182771f9ebbf9402a711.png" alt="=" align="ABSMIDDLE" /> <input name="AnSwEr6" size="10" type="TEXT" value="179/4" /></p>
Explanation / Answer
Alright... to solve the circles center and radius, you must first set it to basic circle form:
The equation is x2+7x+y2-10y-15=0 ...
We must "complete the square" with the x2 and y2:
1) Move the 15 to the other side -> x2+7x+y2-10y = 15
2) Separate the x and y terms -> x2+7x +y2-10y = 15
3) "Complete the square"; add numbers to make the terms squared (don't forget to add to both sides!):
This is a tricky one... remember, to find the square, divide the linear x term by 2,
then square it: Thus,
-add 12.25 to both sides ( (7/2)2 ) to square the x terms,
-add 25 to both sides ( (-10/2)2 ) to square the y terms.
-> x2+7x + 12.25 +y2-10y +25 = 15 + 12.25 + 25
4) Solve the new quadratics : (x+3.5)2 + (y-5)2 = 15 + 12.5 + 25
(x+3.5)2 + (y-5)2 = 52.5
5) Therefore: the center is (-3.5,5) , the radius is 52.5, and the eqn is (x+3.5)2 + (y-5)2 = 52.5
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