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

For external tangents from point R$R$ forming angle 2θ$2\theta$ at the center, the distance d$d$ from center to intersection point obeys θ = (r)/(d)$\cos \theta = \frac{r}{d}$.

Core Logic

Given circle: x² + y² - 4x - 6y - 3 = 0$x^2 + y^2 - 4x - 6y - 3 = 0$ Center O = (2, 3)$O = (2, 3)$ Radius r = √((-2)² + (-3)² - (-3)) = √(4 + 9 + 3) = √(16) = 4$r = \sqrt{(-2)^2 + (-3)^2 - (-3)} = \sqrt{4 + 9 + 3} = \sqrt{16} = 4$.

The tangents PQ and MN intersect at some point R(h, k)$R(h, k)$. The radius vectors OA$OA$ and OB$OB$ subtend angle ∠ AOB = (π)/(3) = 60°$\angle AOB = \frac{\pi}{3} = 60^{\circ}$ at the center. The line joining O$O$ and R$R$ bisects the angle ∠ AOB$\angle AOB$. Thus, ∠ AOR = 30°$\angle AOR = 30^{\circ}$.

Step 1: Apply Trigonometric Relations

In the right-angled triangle Δ AOR$\Delta AOR$, OA$OA$ is the radius (r = 4$r = 4$) and OR$OR$ is the hypotenuse.

(30°) = (OA)/(OR) = (r)/(OR)$$\cos(30^{\circ}) = \frac{OA}{OR} = \frac{r}{OR}$$ √(3)2 = (4)/(OR) ⇒ OR = 8√(3)$$\frac{\sqrt{3}}{2} = \frac{4}{OR} \Rightarrow OR = \frac{8}{\sqrt{3}}$$

Step 2: Construct the Locus Equation

The distance squared between O(2,3)$O(2,3)$ and R(h,k)$R(h,k)$ is OR²$OR^2$:

OR² = (h - 2)² + (k - 3)² = ( 8√(3))²$$OR^2 = (h - 2)^2 + (k - 3)^2 = \left(\frac{8}{\sqrt{3}}\right)^2$$ (h - 2)² + (k - 3)² = (64)/(3)$$(h - 2)^2 + (k - 3)^2 = \frac{64}{3}$$ h² - 4h + 4 + k² - 6k + 9 = (64)/(3)$$h^2 - 4h + 4 + k^2 - 6k + 9 = \frac{64}{3}$$ 3(h² + k² - 4h - 6k + 13) = 64$$3(h^2 + k^2 - 4h - 6k + 13) = 64$$ 3h² + 3k² - 12h - 18k + 39 - 64 = 0$$3h^2 + 3k^2 - 12h - 18k + 39 - 64 = 0$$ 3(h² + k²) - 12h - 18k - 25 = 0$$3(h^2 + k^2) - 12h - 18k - 25 = 0$$

Step 3: Generalize the Equation

Replace (h, k)$(h, k)$ with (x, y)$(x, y)$ for the general locus:

3(x² + y²) - 12x - 18y - 25 = 0$$3(x^2 + y^2) - 12x - 18y - 25 = 0$$

Pattern Recognition

The locus of the intersection of tangents enclosing a constant angle is simply a concentric circle. Its radius expands by 1/ (α/2)$1/\sin(\alpha/2)$ or 1/ (θ)$1/\cos(\theta)$ depending on whether the angle is measured at intersection or center.

Chapter Mix

Class 11 Maths: Conic Sections Class 11 Maths: Straight Lines

Related Formula

Perpendicular bisector properties: m₁m₂ = -1 and mid-point lies on the line.$$\text{Perpendicular bisector properties: } m_1m_2 = -1 \text{ and mid-point lies on the line.}$$ Parametric point on circle x²+y²=r² is (r θ, r θ)$$\text{Parametric point on circle } x^2+y^2=r^2 \text{ is } (r\cos\theta, r\sin\theta)$$

Core Logic

Let P = (2 θ, 2 θ)$P = (2\cos\theta, 2\sin\theta)$. Let Q$Q$ on the line 5x + y + 2 = 0$5x + y + 2 = 0$ be Q(α, -5α-2)$Q(\alpha, -5\alpha-2)$. The line x - y + 1 = 0$x - y + 1 = 0$ is the perpendicular bisector of PQ$PQ$. This gives two conditions: slope of PQ$PQ$ is -1$-1$, and mid-point of PQ$PQ$ satisfies the bisector equation.

Step 1: Apply Slope Condition

Slope of bisector is 1$1$, so slope of PQ$PQ$ must be -1$-1$.

(2 θ - (-5α - 2))/(2 θ - α) = -1$$\frac{2\sin\theta - (-5\alpha - 2)}{2\cos\theta - \alpha} = -1$$ 2 θ + 5α + 2 = -2 θ + α$$2\sin\theta + 5\alpha + 2 = -2\cos\theta + \alpha$$ θ + θ + 2α + 1 = 0 (1)$$\sin\theta + \cos\theta + 2\alpha + 1 = 0 \quad \dots (1)$$

Step 2: Apply Midpoint Condition

Midpoint of PQ$PQ$ is ( (2 θ + α)/(2), (2 θ - 5α - 2)/(2) )$\left( \frac{2\cos\theta + \alpha}{2}, \frac{2\sin\theta - 5\alpha - 2}{2} \right)$. Substitute into x - y + 1 = 0$x - y + 1 = 0$:

(2 θ + α)/(2) - (2 θ - 5α - 2)/(2) + 1 = 0$$\frac{2\cos\theta + \alpha}{2} - \frac{2\sin\theta - 5\alpha - 2}{2} + 1 = 0$$ 2 θ + α - 2 θ + 5α + 2 + 2 = 0$$2\cos\theta + \alpha - 2\sin\theta + 5\alpha + 2 + 2 = 0$$ θ - θ + 3α + 2 = 0 (2)$$\cos\theta - \sin\theta + 3\alpha + 2 = 0 \quad \dots (2)$$

Step 3: Eliminate alpha and Solve

From (1), 2α = - θ - θ - 1 α = (- θ - θ - 1)/(2)$2\alpha = -\sin\theta - \cos\theta - 1 \implies \alpha = \frac{-\sin\theta - \cos\theta - 1}{2}$. Substitute α$\alpha$ into (2):

θ - θ + 3( (- θ - θ - 1)/(2) ) + 2 = 0$$\cos\theta - \sin\theta + 3\left( \frac{-\sin\theta - \cos\theta - 1}{2} \right) + 2 = 0$$ 2 θ - 2 θ - 3 θ - 3 θ - 3 + 4 = 0$$2\cos\theta - 2\sin\theta - 3\sin\theta - 3\cos\theta - 3 + 4 = 0$$ - θ - 5 θ + 1 = 0 θ + 5 θ = 1$$-\cos\theta - 5\sin\theta + 1 = 0 \implies \cos\theta + 5\sin\theta = 1$$

Let's express in half angles:

1 - 2 ²(θ)/(2) + 10 (θ)/(2) (θ)/(2) = 1$$1 - 2\sin^2\frac{\theta}{2} + 10\sin\frac{\theta}{2}\cos\frac{\theta}{2} = 1$$ 2 (θ)/(2) ( 5 (θ)/(2) - (θ)/(2) ) = 0$$2\sin\frac{\theta}{2} \left( 5\cos\frac{\theta}{2} - \sin\frac{\theta}{2} \right) = 0$$

So, (θ)/(2) = 0 θ = 1$\sin\frac{\theta}{2} = 0 \implies \cos\theta = 1$ or (θ)/(2) = 5 θ = (1 - ²(θ/2))/(1 + ²(θ/2)) = (1 - 25)/(1 + 25) = -(24)/(26) = -(12)/(13)$\tan\frac{\theta}{2} = 5 \implies \cos\theta = \frac{1 - \tan^2(\theta/2)}{1 + \tan^2(\theta/2)} = \frac{1 - 25}{1 + 25} = -\frac{24}{26} = -\frac{12}{13}$.

Step 4: Final Calculation

The abscissa of P$P$ is 2 θ$2\cos\theta$. Values of abscissa are 2(1) = 2$2(1) = 2$ and 2(-(12)/(13)) = -(24)/(13)$2\left(-\frac{12}{13}\right) = -\frac{24}{13}$. Sum of abscissa values = 2 - (24)/(13) = (26 - 24)/(13) = (2)/(13)$= 2 - \frac{24}{13} = \frac{26 - 24}{13} = \frac{2}{13}$. We need 13 × (Sum) = 13 × (2)/(13) = 2$13 \times (\text{Sum}) = 13 \times \frac{2}{13} = 2$.

Pattern Recognition

Instead of finding the image of a generic circle point in a line, construct the reflection point Q$Q$ parameter, use slope logic (m₁m₂=-1$m_1m_2=-1$) and midpoint logic simultaneously to create a trigonometric linear equation.

Chapter Mix

Class 11 Maths: Circles Class 11 Maths: Straight Lines

Related Formula

Two circles intersect at distinct points if |r₁ - r₂| < d < r₁ + r₂$$\text{Two circles intersect at distinct points if } |r_1 - r_2| < d < r_1 + r_2$$

where d$d$ is the distance between their centers.

Core Logic

Circle 1: (x + 1)² + (y + 4)² = r²$(x + 1)^2 + (y + 4)^2 = r^2$ Center C₁ = (-1, -4)$C_1 = (-1, -4)$ and Radius r₁ = r$r_1 = r$.

Circle 2: x² + y² - 4x - 2y - 4 = 0 (x - 2)² + (y - 1)² = 3²$x^2 + y^2 - 4x - 2y - 4 = 0 \implies (x - 2)^2 + (y - 1)^2 = 3^2$ Center C₂ = (2, 1)$C_2 = (2, 1)$ and Radius r₂ = 3$r_2 = 3$.

Step 1: Distance Between Centers

Distance d$d$ between C₁$C_1$ and C₂$C_2$:

d = √((2 - (-1))² + (1 - (-4))²)$$d = \sqrt{(2 - (-1))^2 + (1 - (-4))^2}$$ d = √(3² + 5²) = √(9 + 25) = √(34)$$d = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$$

Step 2: Applying the Intersection Condition

For two distinct intersection points:

|r - 3| < √(34) < r + 3$$|r - 3| < \sqrt{34} < r + 3$$

Breaking this down into two inequalities:

• |r - 3| < √(34) -√(34) < r - 3 < √(34) 3 - √(34) < r < 3 + √(34)$|r - 3| < \sqrt{34} \implies -\sqrt{34} < r - 3 < \sqrt{34} \implies 3 - \sqrt{34} < r < 3 + \sqrt{34}$
• r + 3 > √(34) r > √(34) - 3$r + 3 > \sqrt{34} \implies r > \sqrt{34} - 3$

Since radius r > 0$r > 0$, taking the intersection of the conditions:

r in (√(34) - 3, √(34) + 3)$$r \in (\sqrt{34} - 3, \sqrt{34} + 3)$$

Thus, α = √(34) - 3$\alpha = \sqrt{34} - 3$ and β = √(34) + 3$\beta = \sqrt{34} + 3$.

Step 3: Calculating Final Product

αβ = (√(34) - 3)(√(34) + 3) = 34 - 9 = 25$$\alpha\beta = (\sqrt{34} - 3)(\sqrt{34} + 3) = 34 - 9 = 25$$

Pattern Recognition

Intersection of two circles boils down to the fundamental triangle inequality relating the radii to the center distance: |r₁ - r₂| < d < r₁ + r₂$|r_1 - r_2| < d < r_1 + r_2$. Solving this naturally yields an interval (α, β)$(\alpha, \beta)$ formatted as a difference of squares upon multiplication.

Chapter Mix

Class 11 Maths: Circles

Related Formula

The equation of a family of curves passing through the intersection of two conics S₁ = 0$S_1 = 0$ and S₂ = 0$S_2 = 0$ is S₁ + λ S₂ = 0$S_1 + \lambda S_2 = 0$. For this resulting curve to be a circle, the coefficient of x²$x^2$ must equal the coefficient of y²$y^2$, and the coefficient of the xy$xy$ term must be zero.

Core Logic

Let the two ellipses be: S₁ ≡ x² + 2y² - 6x - 12y + 23 = 0$S_1 \equiv x^2 + 2y^2 - 6x - 12y + 23 = 0$ S₂ ≡ 4x² + 2y² - 20x - 12y + 35 = 0$S_2 \equiv 4x^2 + 2y^2 - 20x - 12y + 35 = 0$

Equation of the curve passing through their intersection is S₁ + λ S₂ = 0$S_1 + \lambda S_2 = 0$:

(1 + 4λ)x² + (2 + 2λ)y² - (6 + 20λ)x - (12 + 12λ)y + (23 + 35λ) = 0$$(1 + 4\lambda)x^2 + (2 + 2\lambda)y^2 - (6 + 20\lambda)x - (12 + 12\lambda)y + (23 + 35\lambda) = 0$$

For this to represent a circle, coefficient of x²$x^2$ = coefficient of y²$y^2$:

1 + 4λ = 2 + 2λ 2λ = 1 λ = (1)/(2)$$1 + 4\lambda = 2 + 2\lambda \implies 2\lambda = 1 \implies \lambda = \frac{1}{2}$$

Step 1: Finding Circle Parameters

Substitute λ = 1/2$\lambda = 1/2$ back into the family equation:

(1 + 2)x² + (2 + 1)y² - (6 + 10)x - (12 + 6)y + (23 + (35)/(2)) = 0$$(1 + 2)x^2 + (2 + 1)y^2 - (6 + 10)x - (12 + 6)y + \left(23 + \frac{35}{2}\right) = 0$$ 3x² + 3y² - 16x - 18y + (81)/(2) = 0$$3x^2 + 3y^2 - 16x - 18y + \frac{81}{2} = 0$$

Dividing by 3 to write in standard form:

x² + y² - (16)/(3)x - 6y + (27)/(2) = 0$$x^2 + y^2 - \frac{16}{3}x - 6y + \frac{27}{2} = 0$$

The centre (a, b)$(a, b)$ is given by (-g, -f)$(-g, -f)$:

a = (8)/(3), b = 3$$a = \frac{8}{3}, \quad b = 3$$

The radius r$r$ is given by r² = g² + f² - c$r^2 = g^2 + f^2 - c$:

r² = ((-8)/(3))² + (-3)² - (27)/(2) = (64)/(9) + 9 - (27)/(2) = (128 + 162 - 243)/(18) = (47)/(18)$$r^2 = \left(\frac{-8}{3}\right)^2 + (-3)^2 - \frac{27}{2} = \frac{64}{9} + 9 - \frac{27}{2} = \frac{128 + 162 - 243}{18} = \frac{47}{18}$$

Step 2: Final Calculation

We need to find ab + 18r²$ab + 18r^2$:

ab = ((8)/(3))(3) = 8$$ab = \left(\frac{8}{3}\right)(3) = 8$$ 18r² = 18((47)/(18)) = 47$$18r^2 = 18\left(\frac{47}{18}\right) = 47$$ ab + 18r² = 8 + 47 = 55$$ab + 18r^2 = 8 + 47 = 55$$

Pattern Recognition

When intersection points of two 2nd$2^{\text{nd}}$ degree curves form a circle, apply S₁ + λ S₂ = 0$S_1 + \lambda S_2 = 0$ immediately, forcing the necessary symmetric coefficients to extract λ$\lambda$.

Chapter Mix

Class 11 Maths: Conic Sections Class 11 Maths: Circles

Core Logic

Equation of circle C₂$C_{2}$ with diameter along y=x$y=x$ passing through origin and having length 10. Wait, C₁$C_1$ has diameter 10. The chord y=x$y=x$ passes through (0,0)$(0,0)$. For the chord to be a diameter of C₂$C_2$, the points of intersection with C₁$C_1$ must form the diameter. The center of C₂$C_2$ is the midpoint of the chord. Let the ends of the chord be (0,0)$(0,0)$ and (5,5)$(5,5)$ (since length is √(50)$\sqrt{50}$? Wait, the problem implies the chord of C₁$C_1$ is y=x$y=x$. If C₁$C_1$ is a circle in x ≥ 0$x \geq 0$ of diameter 10 through origin. Center of C₂$C_2$ lies on the chord y=x$y=x$. The equation of circle C₂$C_{2}$ is:

x² + y² - 5x - 5y = 0$$x^2 + y^2 - 5x - 5y = 0$$

Its center is N((5)/(2), (5)/(2))$N\left(\frac{5}{2}, \frac{5}{2}\right)$.

Step 1: Find Farthest Chord

We need the chord of C₂$C_2$ passing through B(2, 3)$B(2, 3)$ which is farthest from the center N((5)/(2), (5)/(2))$N\left(\frac{5}{2}, \frac{5}{2}\right)$. The farthest chord passing through a given point is always perpendicular to the line joining the center to that point. Slope of line NB$NB$:

mNB = (3 - (5)/(2))/(2 - (5)/(2)) = ((1)/(2))/(-(1)/(2)) = -1$$m_{NB} = \frac{3 - \frac{5}{2}}{2 - \frac{5}{2}} = \frac{\frac{1}{2}}{-\frac{1}{2}} = -1$$

Step 2: Chord Equation

The slope of the required chord is perpendicular to mNB$m_{NB}$:

Slope of required chord = 1$$\text{Slope of required chord } = 1$$

Equation of the required chord passing through (2,3)$(2,3)$:

y - 3 = 1(x - 2)$$y - 3 = 1(x - 2)$$

x - y + 1 = 0$x - y + 1 = 0$ Comparing this with x + ay + b = 0$x + ay + b = 0$, we get:

a = -1, b = 1$$a = -1, \quad b = 1$$

Step 3: Final Calculation

a - b = -1 - 1 = -2$$a - b = -1 - 1 = -2$$

Pattern Recognition

The chord of a circle passing through a given internal point that is FARTHEST from the center is exactly the chord that is PERPENDICULAR to the radius (or line segment) connecting the center to that internal point.

Chapter Mix

Class 11 Mathematics: Circles Class 11 Mathematics: Straight Lines