Question

The three vertices of a triangle are (0, 0), (3, 1) and (1, 3). If this triangle is inscribed in a circle, then the equation of the circle is

• A. 2x²+2y²-2x-6y=0
• B. x²+y²-3x-y=0
• C. x²+y²-x-3y=0
• D. 2x²+2y²-6x-2y=0
• E. 2x²+2y²-5x-5y=0

Answer 1

Answer: (E)

The general equation of a circle is given by: \[ x^2 + y^2 + Dx + Ey + F = 0 \] where $D$, $E$, and $F$ are constants. To find these constants, we use the fact that the three points $(0, 0)$, $(3, 1)$, and $(1, 3)$ lie on the circle. Substituting the point $(0, 0)$ into the equation: \[ 0^2 + 0^2 + D(0) + E(0) + F = 0 \quad \Rightarrow \quad F = 0 \] Now, the equation of the circle becomes: \[ x^2 + y^2 + Dx + Ey = 0 \] Substitute the point $(3, 1)$ into this equation: \[ 3^2 + 1^2 + D(3) + E(1) = 0 \quad \Rightarrow \quad 9 + 1 + 3D + E = 0 \quad \Rightarrow \quad 3D + E = -10 \] Substitute the point $(1, 3)$ into the equation: \[ 1^2 + 3^2 + D(1) + E(3) = 0 \quad \Rightarrow \quad 1 + 9 + D + 3E = 0 \quad \Rightarrow \quad D + 3E = -10 \] Now, we solve the system of equations: \[ 3D + E = -10 \quad \text{and} \quad D + 3E = -10 \] Multiplying the second equation by 3: \[ 3D + 9E = -30 \] Subtracting the first equation from this: \[ (3D + 9E) - (3D + E) = -30 + 10 \quad \Rightarrow \quad 8E = -20 \quad \Rightarrow \quad E = -\frac{5}{2} \] Substitute $E = -\frac{5}{2}$ into $3D + E = -10$: \[ 3D - \frac{5}{2} = -10 \quad \Rightarrow \quad 3D = -\frac{15}{2} \quad \Rightarrow \quad D = -\frac{5}{2} \] Thus, the equation of the circle is: \[ x^2 + y^2 - \frac{5}{2}x - \frac{5}{2}y = 0 \] Multiplying the whole equation by 2 to eliminate the fractions: \[ 2x^2 + 2y^2 - 5x - 5y = 0 \]

The correct option is (E) : \(2x^2+2y^2-5x-5y=0\)

Answer 2

Let the equation of the circle be \(x^2 + y^2 + 2gx + 2fy + c = 0\).

Since the circle passes through (0, 0), we have:

\(0^2 + 0^2 + 2g(0) + 2f(0) + c = 0\)

\(c = 0\)

Since the circle passes through (3, 1), we have:

\(3^2 + 1^2 + 2g(3) + 2f(1) = 0\)

\(9 + 1 + 6g + 2f = 0\)

\(6g + 2f = -10\)

\(3g + f = -5\) \((Equation\ 1)\)

Since the circle passes through (1, 3), we have:

\(1^2 + 3^2 + 2g(1) + 2f(3) = 0\)

\(1 + 9 + 2g + 6f = 0\)

\(2g + 6f = -10\)

\(g + 3f = -5\) \((Equation\ 2)\)

Now we solve the system of equations:

\(3g + f = -5\)

\(g + 3f = -5\)

Multiply the second equation by 3: \(3g + 9f = -15\). Subtract the first equation from this result:

\((3g + 9f) - (3g + f) = -15 - (-5)\)

\(8f = -10\)

\(f = -\frac{10}{8} = -\frac{5}{4}\)

Substitute \(f = -\frac{5}{4}\) into the second equation: \(g + 3(-\frac{5}{4}) = -5\), so \(g - \frac{15}{4} = -5\).

\(g = -5 + \frac{15}{4} = \frac{-20 + 15}{4} = -\frac{5}{4}\)

Therefore, the equation of the circle is \(x^2 + y^2 + 2(-\frac{5}{4})x + 2(-\frac{5}{4})y = 0\), which simplifies to:

\(x^2 + y^2 - \frac{5}{2}x - \frac{5}{2}y = 0\)

Multiplying by 2, we get:

\(2x^2 + 2y^2 - 5x - 5y = 0\)