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

To find the constant k such that the circle x^2+y^2-2x+8y+k=0 and the straight line 2x-y+4=0 intersect at only one point, we need to follow these steps:

Step 1: Rewrite the equation of the circle in standard form. The standard form of a circle's equation is (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center of the circle and r is the radius.

The given equation of the circle is x^2+y^2-2x+8y+k=0. Completing the square for x and y terms, we get:

(x-1)^2 + (y+4)^2 = 17 - k

So, the center of the circle is (1, -4) and the radius is sqrt(17 - k).

Step 2: Find the distance from the center of the circle to the line. The distance (d) from a point (x0, y0) to a line Ax + By + C = 0 is given by the formula:

d = |Ax0 + By0 + C| / sqrt(A^2 + B^2)

The equation of the line is 2x - y + 4 = 0. So, A = 2, B = -1, C = 4, and the point is (1, -4). Substituting these values into the formula, we get:

d = |21 - (-1)(-4) + 4| / sqrt(2^2 + (-1)^2) d = |2 - 4 + 4| / sqrt(4 + 1) d = 2 / sqrt(5)

Step 3: 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. So, we set the radius equal to the distance and solve for k:

sqrt(17 - k) = 2 / sqrt(5) Squaring both sides, we get: 17 - k = 4 / 5 Solving for k, we get: k = 17 - 4/5 = 80/5 - 4/5 = 76/5

So, the constant k such that the circle x^2+y^2-2x+8y+k=0 and the straight line 2x-y+4=0 intersect at only one point is 76/5.