Using the standard form of a circle's equation, (x - h)² + (y - k)² = r², and substituting the center (a, -r) and radius r, we get the equation of the circle.

Expanding the equation and simplifying, we can compare it with the given general equation x² + y² - α x + β y + = 0. This comparison allows us to find the values of α, β, and in terms of a and r.

Expanding the equation and simplifying, we can compare it with the given general equation x² + y² - α x + β y + = 0. This comparison allows us to find the values of α, β, and in terms of a and r.

The length of the intercept cut off on the y -axis by the circle x² + y² + 2gx + 2fy + c = 0 is given by 2√(f² - c). From our expanded equation, 2g = -2a, 2f = 2r, and c = a². So, f = r. The y -intercept length is 2√(r² - a²). However, the problem states the intercept length is b. Therefore, b = 2√(r² - a²). This implies b² = 4(r² - a²).

The length of the intercept cut off on the y -axis by the circle x² + y² + 2gx + 2fy + c = 0 is given by 2√(f² - c). From our expanded equation, 2g = -2a, 2f = 2r, and c = a². So, f = r. The y -intercept length is 2√(r² - a²). However, the problem states the intercept length is b. Therefore, b = 2√(r² - a²). This implies b² = 4(r² - a²).

Let the equation of the circle, which touches $x$ -axis at the point $(a, 0)$ , $a > 0$ and cuts off an intercept of length $b$ on $y$ -axis be $x^2 + y^2 - \alpha x + \beta y + \gamma = 0$ . If the circle lies below $x$ -axis, then the ordered pair $(2a, b^2)$ is equal to

Answer: $(2a, 4r^2 - 4a^2)$

Step-by-step solution

Step 1: Determine the center and radius from the tangency condition

Since the circle touches the $x$ -axis at $(a, 0)$ and lies below the $x$ -axis, its center must have coordinates $(a, -r)$ , where $r$ is the radius of the circle. The $y$ -coordinate of the center is negative because the circle is below the $x$ -axis.

Step 2: Write the equation of the circle

Using the standard form of a circle's equation, $(x - h)^2 + (y - k)^2 = r^2$ , and substituting the center $(a, -r)$ and radius $r$ , we get the equation of the circle.

Step 3: Expand and compare with the given equation

Expanding the equation and simplifying, we can compare it with the given general equation $x^2 + y^2 - \alpha x + \beta y + \gamma = 0$ . This comparison allows us to find the values of $\alpha$ , $\beta$ , and $\gamma$ in terms of $a$ and $r$ .

Step 4: Use the y-intercept condition

The length of the intercept cut off on the $y$ -axis by the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $2\sqrt{f^2 - c}$ . From our expanded equation, $2g = -2a$ , $2f = 2r$ , and $c = a^2$ . So, $f = r$ . The $y$ -intercept length is $2\sqrt{r^2 - a^2}$ . However, the problem states the intercept length is $b$ . Therefore, $b = 2\sqrt{r^2 - a^2}$ . This implies $b^2 = 4(r^2 - a^2)$ .

Step 5: Re-evaluate the y-intercept condition based on the problem statement

To find the $y$ -intercept, we set $x=0$ in the circle's equation, resulting in $y^2 + 2ry + a^2 = 0$ . Let the roots of this quadratic equation be $y_1$ and $y_2$ . The length of the intercept $b$ is $|y_1 - y_2|$ . Using Vieta's formulas, $y_1 + y_2 = -2r$ and $y_1y_2 = a^2$ . Substituting these into the formula for the difference of roots, we get $b = 2\sqrt{r^2 - a^2}$ . Squaring both sides gives $b^2 = 4(r^2 - a^2)$ .

Step 6: Solve for the ordered pair

The problem asks for the ordered pair $(2a, b^2)$ . We have already determined $b^2 = 4(r^2 - a^2)$ . Therefore, the ordered pair is $(2a, 4r^2 - 4a^2)$ . The problem statement implies that $b$ is a specific length, which means $r^2 - a^2$ must be positive, so $r > a$ . Without a specific value for $r$ , the expression for $b^2$ will remain in terms of $r$ and $a$ .

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