Question

find the equation of circle which passes through (1,-1) and line 6x+y-18=0 touches the circle at (3,0)

Answer 1

The equation of a circle is:  ( x - a )² + ( y - b )² = r²

The line 6 x + y - 18 = 0 or: y = - 6 x + 18  is a tangent line.

Center of a circle is on the normal which passes through point ( 3, 0 ):

y - y1 = m ( x - x1 ),  where m = 1/6

y - 0 = 1/6 ( x - 3 )  

y = 1/6 x - 1/2 / · 6    becomes  6 y = x - 3     or:   - x + 6 y = - 3

Then the system of equations:

( 1 - a )² + ( - 1 - b )² = r²     and ( 3 - a )² + ( 0 - b )² = r²

1 - 2 a + a² + 1 + 2 b + b² = 9 - 6 a + a² + b²

4 a + 2 b = 7

- a + 6 b = - 3  / · 4

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 4 a + 2 b = 7    ( + )

- 4 a + 24 b = - 12

26 b = - 5,         b = - 5/26

a =  3 - 30/26 = 24/13

r² = 9 - 6 · 24/13 + (24/13)² + (-5/26)² = 925/676

Answer:

The equation is: ( x - 24/13 )² + ( y + 5/26 )² = 925/676