Understanding the Circle Equation and Y-axis Tangency

The given equation of the circle is \(x^2 + y^2 + gx + fy + \frac{e}{4} = 0\). The general equation of a circle is \(x^2 + y^2 + 2Gx + 2Fy + C = 0\), where the center is \((-G, -F)\) and the radius is \(\sqrt{G^2 + F^2 - C}\). Comparing the given equation with the general form, we have:

• \(2G = g \implies G = \frac{g}{2}\)
• \(2F = f \implies F = \frac{f}{2}\)
• \(C = \frac{e}{4}\)

So, the center of the circle is \(\left(-\frac{g}{2}, -\frac{f}{2}\right)\) and the radius is \(r = \sqrt{\left(\frac{g}{2}\right)^2 + \left(\frac{f}{2}\right)^2 - \frac{e}{4}} = \sqrt{\frac{g^2}{4} + \frac{f^2}{4} - \frac{e}{4}}\).

The problem states that the y-axis touches the circle. The equation of the y-axis is \(x = 0\). When a line touches a circle, the distance from the center of the circle to the line is equal to the radius of the circle. The distance of the center \(\left(-\frac{g}{2}, -\frac{f}{2}\right)\) from the line \(x = 0\) is given by the absolute value of the x-coordinate of the center. Distance \(d = \left|-\frac{g}{2}\right| = \left|\frac{g}{2}\right|\).

For the y-axis to touch the circle, the distance \(d\) must equal the radius \(r\): \(d = r\) \(\left|\frac{g}{2}\right| = \sqrt{\frac{g^2}{4} + \frac{f^2}{4} - \frac{e}{4}}\) Squaring both sides: \(\left(\frac{g}{2}\right)^2 = \frac{g^2}{4} + \frac{f^2}{4} - \frac{e}{4}\) \(\frac{g^2}{4} = \frac{g^2}{4} + \frac{f^2}{4} - \frac{e}{4}\) This implies: \(0 = \frac{f^2}{4} - \frac{e}{4}\) Multiplying by 4, we get: \(0 = f^2 - e\) So, the condition for the y-axis to touch the circle is \(e = f^2\).

Finding the Point of Tangency on the Y-axis

The point where the y-axis touches the circle is found by substituting \(x=0\) into the circle equation: \(0^2 + y^2 + g(0) + fy + \frac{e}{4} = 0\) \(y^2 + fy + \frac{e}{4} = 0\) Using the condition \(e = f^2\), substitute \(e\) with \(f^2\): \(y^2 + fy + \frac{f^2}{4} = 0\) This is a perfect square trinomial: \(\left(y + \frac{f}{2}\right)^2 = 0\) The repeated root is \(y = -\frac{f}{2}\).

So, the point of tangency on the y-axis is \(\left(0, -\frac{f}{2}\right)\).

Analyzing the Normal at the Tangent Point

A normal to a circle at any point is the line passing through that point and the center of the circle. The point of tangency is \(P = \left(0, -\frac{f}{2}\right)\). The center of the circle is \(C = \left(-\frac{g}{2}, -\frac{f}{2}\right)\).

The normal line at point P passes through P and C. Let's look at the coordinates of P and C: Point P: \((0, -\frac{f}{2})\) Center C: \((-\frac{g}{2}, -\frac{f}{2})\)

Notice that the y-coordinates of both points are the same \(\left(-\frac{f}{2}\right)\). This means the line passing through P and C is a horizontal line. The equation of the normal line is \(y = -\frac{f}{2}\).

Finding the Intersection of the Normal with the Circle

The normal line \(y = -\frac{f}{2}\) intersects the circle at two points. One point is the point of tangency \(P \left(0, -\frac{f}{2}\right)\). We need to find the other intersection point. Substitute \(y = -\frac{f}{2}\) into the circle equation \(x^2 + y^2 + gx + fy + \frac{e}{4} = 0\): \(x^2 + \left(-\frac{f}{2}\right)^2 + gx + f\left(-\frac{f}{2}\right) + \frac{e}{4} = 0\) \(x^2 + \frac{f^2}{4} + gx - \frac{f^2}{2} + \frac{e}{4} = 0\)

We know that \(e = f^2\). Substitute this into the equation: \(x^2 + \frac{f^2}{4} + gx - \frac{f^2}{2} + \frac{f^2}{4} = 0\) Combine the terms with \(f^2\): \(x^2 + gx + \left(\frac{f^2}{4} - \frac{f^2}{2} + \frac{f^2}{4}\right) = 0\) \(x^2 + gx + \left(\frac{f^2 + f^2 - 2f^2}{4}\right) = 0\) \(x^2 + gx + \left(\frac{2f^2 - 2f^2}{4}\right) = 0\) \(x^2 + gx + 0 = 0\) \(x^2 + gx = 0\) Factor out \(x\): \(x(x + g) = 0\)

This equation gives the x-coordinates of the intersection points. The solutions are \(x=0\) and \(x=-g\).

• If \(x = 0\), then \(y = -\frac{f}{2}\). This is the point of tangency \(\left(0, -\frac{f}{2}\right)\).
• If \(x = -g\), then \(y = -\frac{f}{2}\) (since the intersection points are on the line \(y = -\frac{f}{2}\)). This gives the second intersection point \(\left(-g, -\frac{f}{2}\right)\).

Therefore, the normal at the point of tangency intersects the circle at the point \(\left(-g, -\frac{f}{2}\right)\).

Summary of Points

The coordinates \(\left(-g, -\frac{f}{2}\right)\) match one of the given options.

Revision Table: Circle and Tangency

Additional Information: Geometric Interpretation

When the y-axis touches the circle, the point of tangency is \(\left(0, -\frac{f}{2}\right)\). The center of the circle is \(\left(-\frac{g}{2}, -\frac{f}{2}\right)\). Notice that the y-coordinate of the center and the tangent point are identical. This means the center lies on the horizontal line \(y = -\frac{f}{2}\) which passes through the tangent point.

The normal at the point of tangency is the line segment connecting the tangent point to the center and extending through the center. Since the tangent point and the center have the same y-coordinate, the normal is the horizontal line \(y = -\frac{f}{2}\).

This horizontal normal line \(y = -\frac{f}{2}\) intersects the circle at two points. One point is the tangent point \(\left(0, -\frac{f}{2}\right)\). The other point is on the opposite side of the center along the same horizontal line. The center is at \(x = -\frac{g}{2}\). The tangent point is at \(x=0\). The distance from the tangent point to the center along the x-axis is \(|0 - (-\frac{g}{2})| = |\frac{g}{2}|\). The other intersection point will be the same distance from the center in the opposite x-direction. So, the x-coordinate of the other point is \(-\frac{g}{2} - \left(\frac{g}{2}\right) = -g\), assuming \(g\) is positive (if \(g\) is negative, it's \(-\frac{g}{2} - (-\frac{g}{2}) = 0\), which is the tangent point, and the other point is at \(-\frac{g}{2} + (-\frac{g}{2}) = -g\)). In either case, the x-coordinate is \(-g\). The y-coordinate remains \(-\frac{f}{2}\). Thus, the other intersection point is \(\left(-g, -\frac{f}{2}\right)\). This geometric interpretation confirms the algebraic solution.