SOLUTION: I have a couple of questions on Linear equation. A little bit of explanatiom will help. Thanks a lot. 1. How do I find all points on the y-axis that are 12 units from (5, -3)

Algebra ->  Linear-equations -> SOLUTION: I have a couple of questions on Linear equation. A little bit of explanatiom will help. Thanks a lot. 1. How do I find all points on the y-axis that are 12 units from (5, -3)       Log On


   

Question 40927: I have a couple of questions on Linear equation. A little bit of explanatiom will help. Thanks a lot.
1. How do I find all points on the y-axis that are 12 units from (5, -3)
2. How do I find the equation of a circle that has a diameter with an endpoint (2,3) and ( 4,11). Write the answer in standard form.
Answer by mbarugel(146) About Me  (Show Source):
You can put this solution on YOUR website!
Hello!
All points that are 12 units from (5, -3) form a circle, with center (5, -3) and radius 12. Recall that the formula for a circle with radius r and center (h, k) is:
%28x+-+h%29%5E2+%2B+%28y-k%29%5E2+=+r%5E2
So, in your case:
+%28x-5%29%5E2+%2B+%28y%2B3%29%5E2+=+144
The points on the y-axis are the ones in which x = 0. So using that in our equation:
+%28-5%29%5E2+%2B+%28y%2B3%29%5E2+=+144
+25+%2B+%28y%2B3%29%5E2+=+144
+%28y%2B3%29%5E2+=+119
We get that:
y%5B1%5D+=+sqrt%28119%29-3+=+7.9087
y%5B2%5D+=+-sqrt%28119%29-3+=+-13.9087
Those are the two points on the y-axis that are 12 units away from (5, -3)

I hope this helps!
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