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1:prove that the tangent at the point (1,-2) to the circle x²+y²=5 also touches the circle x²+y²-8x+6y+20=0 .Also find the point of contact.
2:find the length of tangent to the circle
3x²+3y²-7x-6y=12 drawn from (6,-7)​

Respuesta :

mhanifa

#1

Circle 1 has center at (0, 0)

The tangent passes through the center and point (1, -2).

Equation of tangent at (1, -2):

  • x*1 + y*(-2) = 5
  • x - 2y = 5
  • x = 2y + 5

If the same line is tangent to circle 2 then they have one common point.

Solve the system:

  • x = 2y + 5
  • x² + y² - 8x + 6y + 20 = 0

Substitute x:

  • (2y + 5)² + y² - 8(2y + 5) + 6y + 20 = 0
  • 4y² + 20y + 25 - 16y - 40 + 6y + 20 = 0
  • 5y² + 10y + 5 = 0
  • y² + 2y + 1 = 0
  • (y + 1)² = 0
  • y + 1 = 0
  • y = -1 ⇒ x = 2(-1) + 5 = -2 + 5 = 3

We see only one point of contact, (3, -1), this proves the line is tangent to circle 2 as well.

#2

Given circle:

  • 3x² + 3y² - 7x - 6y = 12 and external point (6, -7).

Rewrite the equation of the circle as:

  • 3x² + 3y² - 7x - 6y = 12
  • x² + y² - 7/3x - 2y = 4
  • x² - 2*7/6x + 49/36 + y² - 2y + 1 = 4 + 49/36 + 1
  • (x - 7/6)² + (y - 1)² = 229/36

The center is (7/6, 1) the radius is √229/36

The length of the tangent t is the leg of the right triangle with another leg r being the radius and the hypotenuse d being the distance between the center and the external point:

  • t² = d² - r²
  • t² = (6 - 7/6)² + (-7 - 1)² - 229/36
  • t² = 36 - 14 + 49/36 + 64 - 229/36
  • t² = 86 - 180/36
  • t² = 81
  • t = 9

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