Question

HELP PLEASE (CIRCLES) (40 points)

the line 3y - x = 5 is the tangent to a circle at the point A(1,2). (1) Find the equation of the line passing through A and the centre of the circle. The equation of another line passing through the centre of the circle is x + 3y = 15. (2) Find the equation of the circle.​

Answer 1

• x² + (y - 5)² = 10

Explanation:

Tangent line is:

• 3y - x = 5 ⇒ 3y = x + 5 ⇒ y = 1/3 x + 5/3, the slope is 1/3

Perpendicular line to this has a slope of -3 and passes through point (1, 2).

Its equation is:

• y - 2 = -3(x - 1) ⇒ y = -3x + 3 + 2 ⇒ y = -3x + 5

Using the second line passing through the center, find the coordinates of the center:

• x + 3y = 15 ⇒ 3y = -x + 15 ⇒ y = -1/3x + 5

Solve the system by elimination:

• -3x + 5 = -1/3x + 5 ⇒ x = 0

Then

• y = 5

The center is (0, 5) and radius is the distance from center to point (1, 2):

• r² = (0 - 1)² + (5 - 2)² = 1 + 9 = 10

The equation of circle:

• (x - h)² + (y - k)² = r²
• (x - 0)² + (y - 5)² = 10 ⇒ x² + (y - 5)² = 10