Show that the straight line with equation 3x-y+8=0 is a tangent to the circle with equation x^2+y^2-18x-10y+16=0

Show that the straight line with equation 3xy80 is a tangent to the circle with equation x2y218x10y160 class=

Respuesta :

Answer:

See the works below.

Step-by-step explanation:

One of the way to check if a linear line is tangent to a circle:

Substitute the linear equation to the circle's equation,

  • If the Discriminant (D) = 0, then it is tangent to the circle.
  • If the Discriminant (D) < 0, then it is outside to the circle.
  • If the Discriminant (D) > 0, then it intersects with the circle.

Linear's Equation:

[tex]3x - y + 8 =0[/tex]

[tex]y=3x+8[/tex]

Now, we substitute the y in the Circle's Equation with (3x + 8):

[tex]x^2+y^2-18x-10y+16=0[/tex]

[tex]x^2+(3x+8)^2-18x-10(3x+8)+16=0[/tex]

[tex]x^2+9x^2+48x+64-18x-30x-80+16=0[/tex]

[tex]10x^2=0[/tex]

Thus the coefficients:

  • a = 10
  • b = 0
  • c = 0

[tex]\boxed{Discriminant\ (D)=b^2-4ac}[/tex]

[tex]D=0^2-4(10)(0)[/tex]

   [tex]=0[/tex]

Since the Discriminant = 0, then proven that the Linear Line is tangent to the Circle.