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:

Related Formula

The coordinates for the reflection image of a point (x₁, y₁)$(x_1, y_1)$ across a standard line ax + by + c = 0$ax + by + c = 0$ are determined using:

Core Logic

Find the center and radius of the original given circle:

Reflect the center point (1, -2)$(1, -2)$ across the line mirror 2x - 3y + 5 = 0$2x - 3y + 5 = 0$:

Thus, the center O$O$ of the reflected circle C$C$ is (-3, 4)$(-3, 4)$, and its radius is preserved at r = 3$r = 3$.

Step 1: Locate Point A

We are given that OA$OA$ is ∥$\parallel$ to the x-axis, meaning its y-coordinate matches the center. Since A$A$ lies to the right of the center O(-3, 4)$O(-3, 4)$:

Step 2: Determine Angular Position of Point B

The arc length AB$AB$ is given as (1)/(6)$\frac{1}{6}$ of the total perimeter:

Using parametric coordinates relative to center O(-3, 4)$O(-3, 4)$ with radius r = 3$r = 3$:

Since β < 4$\beta < 4$, the angle θ$\theta$ must point downwards into the negative quadrant relative to A$A$, meaning θ = -60^° = -(π)/(3)$\theta = -60^\circ = -\frac{\pi}{3}$:

Step 3: Evaluate Final Algebraic Value

Substitute the determined coordinates into the target expression:

Pattern Recognition

Whenever parametric configurations on a circle involve radical coordinate multipliers like β - √(3)α$\beta - \sqrt{3}\alpha$, using angular vectors centered at the origin of the circle avoids setting up and solving long distance equations.

Chapter Mix

Class 11 Mathematics: Circles

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Core Logic

The line perpendicular to chord AB$AB$ passing through its midpoint is the diameter of the circle because the perpendicular bisector of any chord passes through the center. Thus, CD$CD$ is a diameter, making its length equal to 2R = 4$2R = 4$.

Step 1: Find Length of Chord AB

The perpendicular distance p$p$ from the center (0,0)$(0,0)$ to the line x + y - 1 = 0$x + y - 1 = 0$ is:

Length of chord AB = 2√(R² - p²) = 2√(4 - (1)/(2)) = 2√((7)/(2)) = √(14)$AB = 2\sqrt{R^2 - p^2} = 2\sqrt{4 - \frac{1}{2}} = 2\sqrt{\frac{7}{2}} = \sqrt{14}$.

Step 2: Area Calculation

Since CD$CD$ is the perpendicular bisector of AB$AB$, the diagonals of quadrilateral ADBC$ADBC$ are perpendicular. Therefore, the area is:

Pattern Recognition

Whenever a line passes through the midpoint of a chord and is perpendicular to it, recognize instantly that it is a diameter. The area of the quadrilateral formed is then simply the area of two triangles sharing the diameter as a base, or (1)/(2) d₁ d₂$\frac{1}{2} d_1 d_2$.

Chapter Mix

Class 11 Mathematics: Circles Class 11 Mathematics: Straight Lines

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This is the equation of a circle passing through diametrically opposite endpoints (x₁, y₁)$(x_1, y_1)$ and (x₂, y₂)$(x_2, y_2)$.

Core Logic

Look at the three known points: A(1, 0)$A(1, 0)$, B(0, 1)$B(0, 1)$, and O(0, 0)$O(0, 0)$. The vectors OA$\vec{OA}$ and OB$\vec{OB}$ align with the x and y axes, meaning ∠ AOB = 90°$\angle AOB = 90^{\circ}$. Since the angle subtended by AB$AB$ at a point O$O$ on the circle is 90°$90^{\circ}$, the segment AB$AB$ must be the diameter of the circle.

Step 1: Finding the Circle Equation

Using the diametric form for points A(1,0)$A(1,0)$ and B(0,1)$B(0,1)$:

Step 2: Solving for k

Since the point (2k, 3k)$(2k, 3k)$ also lies on this circle, substitute x = 2k$x = 2k$ and y = 3k$y = 3k$ into the equation:

13k² - 5k = 0$13k^2 - 5k = 0$ k(13k - 5) = 0$k(13k - 5) = 0$

Step 3: Finding valid points

This gives two possible values for k$k$: k = 0$k = 0$ or k = (5)/(13)$k = \frac{5}{13}$. If k=0$k=0$, the fourth point becomes (0,0)$(0,0)$, which is already listed. The question specifies four distinct points. Thus, k = (5)/(13)$k = \frac{5}{13}$.

Pattern Recognition

Whenever (0,0)$(0,0)$, (a,0)$(a,0)$, and (0,b)$(0,b)$ lie on a circle, they form a right triangle at the origin. The hypotenuse connecting (a,0)$(a,0)$ and (0,b)$(0,b)$ is ALWAYS the diameter. You can immediately write the circle's equation as x² + y² - ax - by = 0$x^2 + y^2 - ax - by = 0$.

Chapter Mix

Class 11 Maths: Circles

Related Formula

The intersection of any two non-parallel diameters yields the center of the circle. A line that intersects a circle at exactly one point is a tangent line. The tangent at the point of contact P$P$ is always perpendicular to the radius CP$CP$, thus mtangent × mradius = -1$m_{\text{tangent}} \times m_{\text{radius}} = -1$.

Core Logic

Find the center C$C$ by solving the two diameter equations:

Multiply (1) by 3 and (2) by 2: 6x - 9y = 15$6x - 9y = 15$ 6x - 8y = 14$6x - 8y = 14$ Subtracting the equations gives -y = 1 ⇒ y = -1$-y = 1 \Rightarrow y = -1$. Substitute y = -1$y = -1$ into (1): 2x + 3 = 5 ⇒ 2x = 2 ⇒ x = 1$2x + 3 = 5 \Rightarrow 2x = 2 \Rightarrow x = 1$. Center C$C$ is (1, -1)$(1, -1)$.

Find the equation of the line joining A(-(22)/(7), -4)$A(-\frac{22}{7}, -4)$ and B(-(1)/(7), 3)$B(-\frac{1}{7}, 3)$. Slope of AB$AB$: mAB = (3 - (-4))/(-1/7 - (-22/7)) = (7)/(21/7) = (7)/(3)$m_{AB} = \frac{3 - (-4)}{-1/7 - (-22/7)} = \frac{7}{21/7} = \frac{7}{3}$. Equation of line AB$AB$:

3y - 9 = 7x + 1$3y - 9 = 7x + 1$

Step 1: Exploit Tangency Geometry

Since line AB$AB$ intersects the circle at only one point P(α, β)$P(\alpha, \beta)$, line AB$AB$ is a tangent to the circle, and P$P$ is the point of tangency. The radius line CP$CP$ is perpendicular to tangent AB$AB$. Slope of CP$CP$ (mCP$m_{CP}$) must be -(3)/(7)$-\frac{3}{7}$.

Equation of the line passing through center C(1, -1)$C(1, -1)$ with slope -(3)/(7)$-\frac{3}{7}$:

Step 2: Solve for Intersection P

Point P(α, β)$P(\alpha, \beta)$ is the intersection of Tangent AB$AB$ and Radius CP$CP$. Solve the system:

49x - 21y = -70$49x - 21y = -70$ 9x + 21y = -12$9x + 21y = -12$ Add them: 58x = -82 ⇒ x = -(82)/(58) = -(41)/(29)$58x = -82 \Rightarrow x = -\frac{82}{58} = -\frac{41}{29}$. So, α = -(41)/(29)$\alpha = -\frac{41}{29}$.

Substitute x$x$ into 3x + 7y = -4$3x + 7y = -4$:

7y = (123 - 116)/(29) = (7)/(29) ⇒ y = (1)/(29)$. So, $β = (1)/(29)$.

Step 3: Evaluate Target Expression

Evaluate $17β - α$:

Pattern Recognition

When a line "intersects a circle at exactly one point", it's a coded cue to stop thinking about quadratics and discriminants, and immediately build a perpendicular geometric radius from the center to find the exact tangency coordinate.

Chapter Mix

Class 11 Mathematics: Circles Class 11 Mathematics: Straight Lines

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Core Logic

Radius of C₁$C_1$ is r₁ = 5$r_1 = 5$ (centered at Origin O(0,0)$O(0,0)$). Radius of C₂$C_2$ is r₂ = 4$r_2 = 4$ (centered at A(α,0)$A(\alpha,0)$ with 5 lt α lt 9$5 \lt \alpha \lt 9$). Let P$P$ be an intersection point. In Δ OAP$\Delta OAP$, the sides are OP = 5$OP = 5$, AP = 4$AP = 4$, and OA = α$OA = \alpha$. The angle between the radii at P$P$ is ∠ OPA = θ$\angle OPA = \theta$, where θ = √(63)8$\sin \theta = \frac{\sqrt{63}}{8}$.

Step 1: Determining Area via the Sine Rule

The area of Δ OAP$\Delta OAP$ can be computed using the two radii and the angle between them:

Step 2: Linking Area to the Common Chord

Step 3: Equating Areas

Equate the two area expressions:

Step 4: Final Evaluation

Square the result as requested:

Pattern Recognition

In intersecting circle problems, the triangle formed by the centers and the intersection point handles both the intersection angle (via cosine/sine area rules) and the common chord (which serves as a perpendicular height doubled).

Chapter Mix

Class 11 Maths: Circles Class 11 Maths: Properties of Triangles