Question

Let \( S = 0 \) be the circle passing through the points \((2, 0)\), \((1, -2)\), and \((-1, 1)\). Then the point \((1, 2)\):

• A. lies inside the circle \( S = 0 \)
• B. lies outside the circle \( S = 0 \)
• C. lies on the circle \( S = 0 \)
• D. is the centre of the circle \( S = 0 \)

Hint:

To determine a point's position relative to a circle \( x^2 + y^2 + Dx + Ey + F = 0 \), substitute the point into the equation: if \( S>0 \), the point is outside; if \( S = 0 \), on; if \( S<0 \), inside.

Answer 1

The Correct Option is B

Step 1: Write the general equation of the circle.
The equation of a circle is: \[ x^2 + y^2 + Dx + Ey + F = 0 \] We need to find \( D \), \( E \), and \( F \) using the given points.
Step 2: Substitute the points to form equations.
For \((2, 0)\): \[ 4 + 2D + F = 0 \implies 2D + F = -4 \quad (1) \] For \((1, -2)\): \[ 1 + 4 + D - 2E + F = 0 \implies D - 2E + F = -5 \quad (2) \] For \((-1, 1)\): \[ 1 + 1 - D + E + F = 0 \implies -D + E + F = -2 \quad (3) \]
Step 3: Solve the system of equations.
Subtract (1) from (2): \[ (D - 2E + F) - (2D + F) = -5 - (-4) \implies -D - 2E = -1 \implies D + 2E = 1 \quad (4) \] Subtract (3) from (2): \[ (D - 2E + F) - (-D + E + F) = -5 - (-2) \implies 2D - 3E = -3 \quad (5) \] Solve (4) and (5): From (4), \( D + 2E = 1 \). Multiply by 2: \( 2D + 4E = 2 \). Subtract (5): \[ (2D + 4E) - (2D - 3E) = 2 - (-3) \implies 7E = 5 \implies E = \frac{5}{7} \] Then, \( D + 2\left(\frac{5}{7}\right) = 1 \implies D = -\frac{3}{7} \). Substitute into (1): \[ 2\left(-\frac{3}{7}\right) + F = -4 \implies F = -\frac{22}{7} \] The equation is: \[ x^2 + y^2 - \frac{3}{7}x + \frac{5}{7}y - \frac{22}{7} = 0 \quad \text{or} \quad 7x^2 + 7y^2 - 3x + 5y - 22 = 0 \]
Step 4: Determine the position of \((1, 2)\).
Substitute \((1, 2)\): \[ S(1, 2) = 1 + 4 - \frac{3}{7} + \frac{10}{7} - \frac{22}{7} = \frac{20}{7} \] Since \( S(1, 2)>0 \), the point lies outside the circle.
Step 5: Check if \((1, 2)\) is the center.
Center: \( \left(-\frac{D}{2}, -\frac{E}{2}\right) = \left(\frac{3}{14}, -\frac{5}{14}\right) \neq (1, 2) \). Final Answer: \[ \boxed{2} \]