Point (2,3) is on a circle with the center (3,5). What is the radius of the circle?
2 answers:
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let's calculate it step by step.
I'll assume that the function we look for is quadratic, because we only got 3 points
I plotted the 3 points for checking the equation later, and some intuited insight. see screenshot 1
let's write the general form of quadratic equations:
f(x) = a*x² + b*x + c
our goal is to correctly determine a, b and c
since we know from the 2nd point that
f(0) = a*0² + b*0 + c = 4
we can say
<em><u>c = 4</u></em>
this one comes almost for free
now to the next point, the 3rd one
f(1) = a*1² + b*1 + 4 = -3
we can write it as
a + b + 4 = -3
we can rearrange that to
<em><u>a = -b -7</u></em>
lastly, to the first point
f(-2) = a*(-2)² + b*(-2) + 4 = 5
simplified to
4a -2b + 4 = 5
since sort of defined a, let's plug it into our newest equation
4*(-b-7) -2b +4 = 5
we just swapped a for what its equal to.
now we just got numbers and b's
simplify it to
-4b -28 -2b +4 = 5
-6b -24 = 5
-6b = 29
<em><u>b = -29/6</u></em>
now we use the exact value of b to plug it into the rather simple equation from earlier. into this one
a = -b -7
a = -(-29/6) -7
a = 29/6 -42/6
<em><u>a= -13/6</u></em>
now that we got a, b and c we can write the whole equation
<em><u>f(x) = -13/6*x² -29/6*x +4</u></em>
see the graphic check in screenshot 2
second part of the problem:
if we want to move the whole graph 4 units down, we effectivly just cancel the +4 part
g(x) = f(x) -4 = f(x) = -13/6*x² -29/6*x +4 -4
<em><u>g(x) = -13/6*x² -29/6*x</u></em>
see screenshot 3 for the graph
