Let the equation of the circle, which touches x-axis at the point (a, 0), a > 0$a > 0$ and cuts off an intercept of length b on y-axis be x² + y² - α x + β y + γ = 0$x^2 + y^2 - \alpha x + \beta y + \gamma = 0$. If the circle lies below x-axis, then the ordered pair (2a, b²)$(2a, b^2)$ is equal to:

Core Logic

Intersection of circle with x-axis provides A(2,0)$A(2,0)$ and B(-2,0)$B(-2,0)$. Let the point of intersection of AQ$AQ$ and BP$BP$ be R(h, k)$R(h, k)$. Since R$R$ lies on BP$BP$, the slope mBR = mBP$m_{BR} = m_{BP}$:

Since R$R$ lies on AQ$AQ$, the slope mAR = mAQ$m_{AR} = m_{AQ}$:

Execution

We are given α - β = (π)/(2) ⇒ (α)/(2) - (β)/(2) = (π)/(4)$\alpha - \beta = \frac{\pi}{2} \Rightarrow \frac{\alpha}{2} - \frac{\beta}{2} = \frac{\pi}{4}$. Applying the (A-B)$\tan(A-B)$ formula:

Substitute the slope relations: (α)/(2) = (k)/(h+2)$\tan\frac{\alpha}{2} = \frac{k}{h+2}$ (β)/(2) = -(h-2)/(k)$\tan\frac{\beta}{2} = -\frac{h-2}{k}$ (since - (β)/(2) = (k)/(h-2)$-\cot\frac{\beta}{2} = \frac{k}{h-2}$)

1 = ((k)/(h+2) + (h-2)/(k))/(1 + ((k)/(h+2))((2-h)/(k)))$1 = \frac{\frac{k}{h+2} + \frac{h-2}{k}}{1 + \left(\frac{k}{h+2}\right)\left(\frac{2-h}{k}\right)}$ 1 = k² + h² - 4(k(h+2))/(k) · (k(h+2) - k(2-h))/(k(h+2)) wait, clear denominator$1 = \frac{k^2 + h^2 - 4}{\frac{k(h+2)}{k} \cdot \frac{k(h+2) - k(2-h)}{k(h+2)} \dots \text{ wait, clear denominator}}$ 1 = (k² + h² - 4)/(k(h+2) + k(2-h)) × k(h+2)$1 = \frac{k^2 + h^2 - 4}{k(h+2) + k(2-h)} \times k(h+2) \dots$ The denominator simplifies to:

Numerator is (k² + h² - 4)/(k(h+2))$\frac{k^2 + h^2 - 4}{k(h+2)}$. So the expression simplifies to:

Locus of R$R$ is x² + y² - 4y - 4 = 0$x^2 + y^2 - 4y - 4 = 0$.

Pattern Recognition

Connecting chords from extreme diameter vertices to points whose parametric angles differ by π/2$\pi/2$ reliably generates perpendicular-like slope products or standard tangent angle identities, mapping directly to a circular locus.

Chapter Mix

Class 11 Maths: Circles

Related Formula

Perpendicular distance from point (x₀, y₀)$(x_0, y_0)$ to line Ax + By + C = 0$Ax + By + C = 0$:

Core Logic

Since the circle touches two symmetric intersecting lines, its center must lie on their angle bisector (x$x$-axis). Use this property to find the center parameters.

Step 1: Establish Center and Radius Equations

The lines are x+y-3=0$x+y-3=0$ and x-y-3=0$x-y-3=0$. The intersection point is (3,0)$(3,0)$, and the bisector line is the x$x$-axis. Let the center be C(a, 0)$C(a, 0)$. The radius r$r$ is the perpendicular distance to either line:

Step 2: Apply Point Passage Constraint

The circle equation is (x - a)² + y² = r²$(x - a)^2 + y^2 = r^2$. Substitute the given passage point (-9, 4)$(-9, 4)$:

Step 3: Solve for Quadratic Roots

Factor the quadratic equation:

Step 4: Compute Radii Squares Difference

Find the corresponding radius value for each root:

The absolute difference between their squares is:

Pattern Recognition

Recognizing that the center must lie on the line of symmetry (x$x$-axis) eliminates one variable parameter immediately, reducing a difficult geometric system to a simple single-variable quadratic equation.

Chapter Mix

Class 11 Mathematics: Circles Class 11 Mathematics: Straight Lines

Related Formula

For a circle in the third quadrant touching both coordinate axes, the center layout is (-r, -r)$(-r, -r)$ and equation looks like:

For external contact between circles C₁$C_1$ and C₂$C_2$, the distance between centers equals the sum of their radii:

Core Logic

Circle C₁$C_1$ has radius r₁ = 3$r_1 = 3$ and touches both axes in the third quadrant, so its center is A(-3, -3)$A(-3, -3)$. Circle C₂$C_2$ has center B(1, 3)$B(1, 3)$.

The distance between centers A$A$ and B$B$ is:

Step 1: Determine Radius of Circle 2

Step 2: Locate the Contact Point via Section Formula

The point of contact P(α, β)$P(\alpha, \beta)$ divides the line segment joining centers A(-3, -3)$A(-3, -3)$ and B(1, 3)$B(1, 3)$ internally in the ratio r₁ : r₂ = 3 : (2√(13) - 3)$r_1 : r_2 = 3 : (2\sqrt{13} - 3)$.

Using the internal section formula:

Step 3: Calculate the Difference Value

Find (β - α)²$(\beta - \alpha)^2$:

Comparing with (m)/(n)$\frac{m}{n}$ where (m, n) = 1$\gcd(m, n) = 1$ gives m = 9, n = 13$m = 9, n = 13$.

Pattern Recognition

Notice that computing (β - α)$(\beta - \alpha)$ directly cancels out the irrational √(13)$\sqrt{13}$ term from the numerator before squaring, saving a significant amount of tedious arithmetic expansion.

Chapter Mix

Class 11 Mathematics: Coordinate Geometry Class 11 Mathematics: Circles

Related Formula

Length of a chord with perpendicular distance d$d$ from the center of a circle of radius r$r$ is:

Core Logic

Points A(4,2)$A(4,2)$ and B(0,2)$B(0,2)$ have the same y-coordinate, meaning chord AB$AB$ is horizontal. The perpendicular bisector of a horizontal chord is vertical.

Midpoint of AB$AB$ is M(2,2)$M(2,2)$. Thus, the vertical line passing through the center is x = 2$x = 2$.

Step 1: Identify Center and Radius

Since the center lies on the line 3x + 2y + 2 = 0$3x + 2y + 2 = 0$, substitute x = 2$x = 2$ to find the y-coordinate:

So, Center O = (2, -4)$O = (2, -4)$.

Calculate radius r$r$ using point B(0,2)$B(0,2)$:

Step 2: Find Target Chord Length

The targeted chord has a given midpoint N(1,2)$N(1,2)$. Distance from center O(2,-4)$O(2,-4)$ to N(1,2)$N(1,2)$:

Pattern Recognition

Points sharing a coordinate define standard vertical or horizontal perpendicular configurations immediately. Always exploit geometrical configurations before jumping into standard circle equations.

Chapter Mix

Class 11 Mathematics: Circles