SOLUTION: Find the equation of the circle if it is tangent to the line x+y=2 at point (4,-2) and the center at x-axis.

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Question 678416: Find the equation of the circle if it is tangent to the line x+y=2 at point (4,-2) and the center at x-axis.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

%28x+-+h%29%5E+2+%2B+%28y+-+k%29+%5E2+=+r%5E2
(h,k)-center and r radius
given:
given:
the circle if it is tangent to the line x%2By=2 at point (4,-2) and the center at x-axis

find the equation of the circle:
Let ... C(a,0) be the center.
Also, let the line L :x%2By=2 touch
this circle at the point A(4,-2)
Line x+%2B+y+=+2 y+=+-%281%29x+%2B2 has slope m%5B1%5D+=+-1
and slope of seg AC is m%5B2%5D+=+%28+-2-0+%29+%2F+%28+4+-a+%29+=+-2+%2F+%28+4-a+%29.
Since line L is perpendicular to segment AC, +m%5B1%5D%2Am%5B2+%5D=+1
%28-1%29+%2A+%28-2+%2F+%28+4+-a+%29+%29+=+1
-2%2F+%28+4+-a+%29+=+1
-2=++4-a+
a+=+4%2B2
a+=+6
Hence, C ≡ (a,0) ≡ (6,0)
and
r%5E2+=+%28AC%29%5E2+=+%286-4%29%5E2+%2B+%280%2B2%29%5E2+=%282%29%5E2+%2B+%282%29%5E2+=+8.
Hence, by Center-Radius Form, eq of circle is:
%28+x+-+h+%29%5E2+%2B+%28+y+-+k+%29%5E2++=+r%5E2+
%28+x+-+2+%29%5E2++%2B+%28+y+-+0+%29%5E2++=+8+

%28+x+-+2+%29%5E2++%2B+y+%5E2++=+8......... your answer