所選文字: find the constant k such that the circle x^2+y^2-2x+8y+k=0 and the straight line2x-y+4=0 intersact at only one point 回答問題

The equation of the circle can be rewritten in standard form as (x-1)^2 + (y+4)^2 = r^2, where r is the radius of the circle. The constant k is related to the radius by the equation k = -r^2 - 17.

The distance d from the center of the circle (1, -4) to the line 2x - y + 4 = 0 can be found using the formula d = |Ax1 + By1 + C| / sqrt(A^2 + B^2), where (x1, y1) is the center of the circle and Ax + By + C = 0 is the equation of the line. Substituting the given values, we get d = |2*1 - (-4) + 4| / sqrt(2^2 + (-1)^2) = 6 / sqrt(5).

For the circle and the line to intersect at only one point, the distance from the center of the circle to the line must be equal to the radius of the circle. Therefore, we have r = d = 6 / sqrt(5).

Substituting r into the equation for k, we get k = -(6 / sqrt(5))^2 - 17 = -72/5 - 17 = -157/5 = -31.4.