Question

For each parabola \( y = x^2 + px + q \), meeting the coordinate axes at three distinct points, if circles are drawn through these points, then the family of circles must pass through:

• A. \( (1, 0) \)
• B. \( (0, 1) \)
• C. \( (1, 1) \)
• D. \( (p, q) \)

Hint:

When dealing with curves like parabolas and families of circles passing through specific points, the geometry often dictates that the family of circles passes through a fixed point, which can be found through intersection properties.

Answer 1

The Correct Option is B

Suppose the parabola is given by \(y = x^2 + px + q\) and it cuts the \(y\)-axis at points \(A(\alpha, 0)\) and \(B(\beta, 0)\). Then, \(\alpha\) and \(\beta\) are roots of the equation: \[ x^2 + px + q = 0 \] Therefore, \(\alpha + \beta = -p\) and \(\alpha\beta = q\). The parabola \(y = x^2 + px + q\) cuts the \(y\)-axis at \((0, q)\). Let the equation of the circle passing through \(A\), \(B\), and \(C\) be: \[ x^2 + y^2 + 2gx + 2fy + c = 0 \quad {(i)} \] From points \(A\) and \(B\): \[ \alpha^2 + 2g\alpha + c = 0 \quad {(ii)} \] \[ \beta^2 + 2g\beta + c = 0 \quad {(iii)} \] And for point \(C(0, q)\): \[ q^2 + 2fq + c = 0 \quad {(iv)} \] Subtracting equation (iii) from (ii), we have: \[ \alpha + \beta + 2g = 0 \quad \Rightarrow \quad g = -\frac{p}{2} \] Adding equations (ii) and (iii), we get: \[ \alpha^2 + \beta^2 + 2g(\alpha + \beta) + 2c = 0 \] \[ (\alpha + \beta)^2 - 2\alpha\beta + 2g(\alpha + \beta) + 2c = 0 \] \[ p^2 - 2q - p^2 + 2c = 0 \quad \Rightarrow \quad c = q - \frac{p^2}{2} \] Using \(c = q\) in equation (iv), we solve: \[ q^2 + 2fq + q = 0 \quad \Rightarrow \quad f = -\left(q + \frac{1}{2}\right) \] Substituting the values of \(g\), \(f\), and \(c\) back into equation (i), we derive the equation of the family of circles passing through points \(A\), \(B\), and \(C\): \[ x^2 + y^2 - px + (q + 1)y + q = 0 \] Clearly, it passes through \((0, 1)\). Thus, the correct answer is Option B.