find the equation of the circle passing through the points (2 3) and (-1 1). and touches the line x – 3y – 11 = 0

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The work is simple but tedious. The equation of the circle is (x - h)² + (y - k)² = R², where
h = (-239 + 40√39)/6 ≈ 1.7999867
k = ( 125 - 20√39)/2 ≈ 0.0500200
R² = (393250 - 62920√39)/36 ≈ 8.7423872

I will explain how I arrived at the answers. Let me cross check and organise my rough work.

SOLUTION

(1) The standard form of the equation of a circle is (x - h)² + (y - k)² = R². Since the circle passes through points (2, 3) and (-1, 1), these coordinates must satisfy the equation of the circle in question. Therefore,
(2 - h)² + (3 - k)² = R² ------- (1)
(-1 - h)² + (1 - k)² = R² ----- (2)
Expanding leads to
h² + k² - 6k - 4h + 13 = R² ----- (1)
h² + k² - 2k + 2h + 2 = R² ----- (2)

(2) Let (x', y') be the point at which the circle touches the line x - 3y - 11 = 0. It follows that the slope of the circle at (x', y') is ⅓ just as the slope of the line is ⅓. After all the line x - 3y - 11 = 0 is a tangent line to the circle at (x', y'). The next step is to find (x', y').

3) Given that the slope of the general circle is dy/dx = -(x - h)/(y - k). The point (x', y') simultaneously satisfies the equations:
-(x - h)/(y - k) = ⅓ --------- (a)
x - 3y - 11 = 0 ----------- (b)

Eq.(a) can be rewritten as -3x + 3h + k = y. Solving for y in eq.(b) and substituting into eq.(a) leads to the following solution.
x' = (9h + 3k + 11)/10 , y' = (3h + k - 33)/10

Since (x', y') is a point on the circle, it must also satisfy the equation of the circle. Substituting (x', y') into the circle's equation leads to:
( h² + 9k² - 22h + 66k - 6hk + 121)/10 = R² -------- (3)

4) You have to solve the 3 equations simultaneously :
h² + k² - 6k - 4h + 13 = R² ----- (1)
h² + k² - 2k + 2h + 2 = R² ----- (2)
(h² + 9k² - 22h + 66k - 6hk + 121)/10 = R² -------- (3)

5) Subtracting eq.(1) from eq.(2) yields
4k + 6h - 11 = 0. ------- (4)

Subtracting eq.(2) from eq.(3) yields
9h² + k² - 86k + 42h + 60hk - 101 = 0 ------ (5)

Solving for h in eq.(4) and substituting into eq.(5) yields :
k² - 125k + (25/4) = 0

Solving this quadratic equation yields the solution
k = (125 ± 20√39)/2 and substituting into 4k + 6h - 11 = 0 yields the following solution for h :
h = -(239 ± 40√39)/6

6) Making a sketch of the points (2, 3) & (-1, 1) and the line x - 3y - 11 = 0, we can conclude that h = (-239 + 40√39)/6 ≈ 1.7999867 and k = (125 - 20√39)/2 ≈ 0.0500200. The centre of the circle is ( (-239 + 40√39)/6, (125 - 20√39)/2 ) . Substitute the values for h & k into either eq.(1) or eq.(2) yields the value for the radius :
R² = (393250 - 62920√39)/36 ≈ 8.7423872
R ≈ 2.95675281

7) The equation of the circle passing through (2, 3) & (-1, 1) and touching the line x - 3y - 11 = 0 is

(x - h)² + (y - k)² = R², where

h = (-239 + 40√39)/6 ≈ 1.7999867

k = ( 125 - 20√39)/2 ≈ 0.0500200

R² = (393250 - 62920√39)/36 ≈ 8.7423872

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