<PRE><B>Question 738620</B>
The standard equation for a line is {{{(x-h)^2 + (y-k)^2 = r^2}}} where (h.k) is the center and r is the radius.&lt;P&gt;
(4,1) lies on the circle, so {{{(4-h)^2 + (1-k)^2 = r^2}}}  Call that eq: 1.&lt;P&gt;
(6,5) lies on the circle, so {{{(6-h)^2 + (5-k)^2 = r^2}}}   Call that eq: 2&lt;P&gt;
(h,k) is the center of the circle, and therefore must lie on the line that passes through the center.  4h + k = 16 and thus k = 16-4h  eq: 3&lt;P&gt;
Set 1 = 2 since both = r squared.  {{{(4-h)^2 + (1-k)^2 =(6-h)^2 + (5-k)^2}}}&lt;P&gt;
Expand the squared binomials:  {{{16 - 8h + h^2 + 1 -2k + k^2 = 36 - 12h + h^2 + 25 -10k + k^2}}}.&lt;P&gt;
Subtract h squared and k squared from both sides, and combine like terms.&lt;P&gt;
17 - 8h - 2k = 61 -12h - 10k&lt;P&gt;
4h = 44 - 8k so h = 11-2k  Substitute that into eq:3 and solve it for k&lt;P&gt;
k = 16 - 4(11-2k) = 16 - 44 +8k = 28+8k&lt;P&gt;
-7k =28&lt;P&gt;
k = -4&lt;P&gt;
h = 11-2k = 11-2(-4) = 11+8 = 19.&lt;P&gt;
The center of the circle is (19,-4) and thus the equation is:&lt;P&gt;
{{{(x-19)^2 + (y-(-4))^2 = r^2}}} &lt;P&gt;
Use one of the given points to find r.  Use (4,1).&lt;P&gt;
{{{(4-19)^2 + (1+4)^2 = r^2 = (-15)^2 +5^2 = 225 + 25 = 250}}}
The equation is {{{(x-19)^2 + (y-(-4))^2 = 250}}}
&lt;P&gt;Hope the solution helped.  Sometimes you need more than a solution.  Contact fcabanski@hotmail.com for online, private tutoring, or personalized problem solving (quick for groups of problems.)

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