standard equation of circle concentric with X^2+Y^2-2X-8Y+1=0 and tangent to line 2X-Y=3

Question

Complete the squares to find the center of the circle.
X^2+Y^2-2X-8Y+1=0
(x-1)^2+(y-4)^2 - 16 =0
The center is (1,4) and the radius for the given circle is 4.
Now find the distance from (1,4) to the line to get the radius, then write the concentric circle's equation.

how do I get the distance if I don't have a point of intersection with the line or do I?

If you have a point (x1,y1), the distance to the line Ax+By+C=0 is
|Ax1+By1+c|/sqrt(A^2+B^2)
No, you don't have a point of intersection, you just want to find the distance and use that for the radius. The circle and line need to be tangent.

Thank you so much...

Answer 1

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