A secant, because the system has two solutions and the center of the circle does not lie on the line.
Practice makes perfect
We are given the equations of the circle and the line, and we want to determine whether the line is a tangent, a secant, a secant that contains a diameter, or none of these.
Circle:& (x-5)^2+(y+1)^2=4
Line:& y=1/5x-3
To classify the line we should find the number of points of intersection of the line and the circle. To do this we will solve the system of equations using the Substitution Method.
Next we will use the Quadratic Formula to solve the first equation.
ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a
We first need to identify the values of a, b, and c.
26x^2-270x+625=0 ⇕ 26x^2+( -270)x+ 625=0
We see that a= 26, b= -270, and c= 625. Let's substitute these values into the Quadratic Formula.
The solutions for this equation are x= 270± sqrt(7900)52. Let's separate them into the positive and negative cases.
x=270± sqrt(7900)/52
x_1=270+sqrt(7900)/52
x_2=270-sqrt(7900)/52
x_1=270+88.88.../52
x_2=270-88.88.../52
x_1≈ 6.9
x_2≈ 3.5
Now we can find corresponding y-values using the second equation.
y=1/5x-3
y_1≈1/5( 6.9)-3
y_2≈1/5( 3.5)-3
y_1≈1.38-3
y_2≈0.7-3
y_1≈ -1.62
y_2≈ -2.3
We found two points of intersection of the circle and the line. This means that the line is the secant. However, we should check whether this secant contains a diameter. To do this we can check if the center of the circle lies on the line. Let's rewrite the circle equation to identify the center.
(x-5)^2+(y+1)^2=4 ⇓
(x-5)^2+(y-(-1))^2=4
The center of this circle occurs at (5,-1). Now we will substitute this point into the line equation and see if we end with a true statement.
Since we end with a false statement, the center of the circle does not lie on the line. Therefore the line is a secant that does not contain a diameter.
