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Prove algebraically that the straight line with equation x-2y=10 is a tangent to the circle with equation x^2 + y^2 = 20

User Kaoskeya
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Answer:

See below

Explanation:

We solve the equations as a system of simultaneous equations to find the point of intersection. The point of intersection is the point of tangency.

x - 2y = 10

x² + y² = 20

x = 2y + 10

(2y + 10)² + y² = 20

x = 2y + 10

4y² + 40y + 100 + y² = 20

x = 2y + 10

5y² + 40y + 80 = 0

x = 2y + 10

y² + 8y + 16 = 0

x = 2y + 10

(y + 4)² = 0

x = 2(-4) + 10

y =-4

x = 2

y = -4

The point of tangency is (2, -4).

The equation of the circle is x² + y² = 0. The center of the circle is (0, 0). The point of tangency is (2, -4). The line in which the radius to the point of tangency lies has slope

m = (-4 - 0)/(2 - 0) = -2

The given line has equation

x - 2y = 10

-2y = -x + 10

y = (1/2)x - 5,

so its slope is 1/2.

The radius of a circle drawn to the point of tangency and the tangent are perpendicular.

The slopes are -2 and 1/2, respectively, which are negative reciprocals, so they are perpendicular. Therefore, line x - 2y = 10 is a tangent to circle x² + y² = 20.

User Ken Barber
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