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

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...

just believe in god

You're welcome! I'm glad I could help. Remember, if you need any more assistance or have any more questions, feel free to ask! I'm here to entertain and assist you.

You're welcome! If you have any more questions, feel free to ask.

You're welcome! To find the distance from the center of the circle (1,4) to the line 2X-Y=3, you can use the formula I mentioned earlier. Since the coefficients of X and Y in the line equation are 2 and -1 respectively, and the constant term is 3, we have A=2, B=-1, and C=-3.

Using the formula |Ax1+By1+C|/sqrt(A^2+B^2), we can substitute the values to find the distance. Plugging in x1=1, y1=4, A=2, B=-1, and C=-3, we get

|2(1) + (-1)(4) - 3| / sqrt(2^2 + (-1)^2)
|2 - 4 - 3| / sqrt(4 + 1)
|-5| / sqrt(5)
5 / sqrt(5)

Simplifying further, we get

5 / sqrt(5) = sqrt(5)

Therefore, the distance from the center of the circle to the line is sqrt(5), which will be the radius of the concentric circle.

Now you can use this radius and the center (1,4) to write the equation for the concentric circle.