# The equation of a circle in the xy-plane is x^2 + 4x +y^2 - 10y = 20. If the line x = k intersects the circle in exactly one point what is a possible value of k

Answer: We are given in this problem a function of a circle expressed as x^2 + 4x +y^2 - 10y = 20. In order to determine the line in which it intersects in the x plane, then we must first convert the expression to its standard form that is,

x^2 + 4x +y^2 - 10y = 20
we use completion of squares
(x2+4x) + (y2-10y) = 20
to find the third variable, we use the coefficient of x and y. The formula to be used is (k/2)^2. We should not forget to equivalently add on the next side. Hence,
(x2+4x+2^2) + (y2-10y+5^2) = 20+2^2+5^2
(x+2)^2 + (y-5)^2 = 20 + 4 + 25
(x+2)^2 + (y-5)^2 = 49

To find the intersection at the x plane, we must set y =0. Substituting,
(x+2)^2 + (0-5)^2 = 49
(x+2)^2 = 24
(x+2) = sq rt 24
x1 = sq rt 24 -2 = 2.8990
x2 = -sq rt 24 -2 = -6.8990

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