Let the circle x^2 + y^2 = 4$x^{2} + y^{2} = 4$ intersect x-axis at the points A(a, 0), a > 0$A(a, 0), a > 0$ and B(b, 0)$B(b, 0)$. Let P(2 cosalpha, 2 sinalpha), 0 < alpha < fracpi2$P(2 \cos\alpha, 2 \sin\alpha), 0 < \alpha < \frac{\pi}{2}$ and Q(2 cosbeta, 2 sinbeta)$Q(2 \cos\beta, 2 \sin\beta)$ be two points such that (alpha - beta) = fracpi2$(\alpha - \beta) = \frac{\pi}{2}$. Then the point of intersection of AQ$AQ$ and BP$BP$ lies on:

### Related Formula For external tangents from point R$R$ forming angle 2theta$2\theta$ at the center, the distance d$d$ from center to intersection point obeys cos theta = fracrd$\cos \theta = \frac{r}{d}$. ### Core Logic Given circle: x^2 + y^2 - 4x - 6y - 3 = 0$x^2 + y^2 - 4x - 6y - 3 = 0$ Center O = (2, 3)$O = (2, 3)$ Radius r = sqrt(-2)^2 + (-3)^2 - (-3) = sqrt4 + 9 + 3 = sqrt16 = 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 angle AOB = fracpi3 = 60^circ$\angle AOB = \frac{\pi}{3} = 60^{\circ}$ at the center. The line joining O$O$ and R$R$ bisects the angle angle AOB$\angle AOB$. Thus, angle AOR = 30^circ$\angle AOR = 30^{\circ}$. ### Step 1: Apply Trigonometric Relations

In the right-angled triangle Delta AOR$\Delta AOR$, OA$OA$ is the radius (r = 4$r = 4$) and OR$OR$ is the hypotenuse. cos(30^circ) = fracOAOR = fracrOR$$\cos(30^{\circ}) = \frac{OA}{OR} = \frac{r}{OR}$$ fracsqrt32 = frac4OR Rightarrow OR = frac8sqrt3$$\frac{\sqrt{3}}{2} = \frac{4}{OR} \Rightarrow OR = \frac{8}{\sqrt{3}}$$ ### Step 2: Construct the Locus Equation The distance squared between O(2,3)$O(2,3)$ and R(h,k)$R(h,k)$ is OR^2$OR^2$: OR^2 = (h - 2)^2 + (k - 3)^2 = left(frac8sqrt3right)^2$$OR^2 = (h - 2)^2 + (k - 3)^2 = \left(\frac{8}{\sqrt{3}}\right)^2$$ (h - 2)^2 + (k - 3)^2 = frac643$$(h - 2)^2 + (k - 3)^2 = \frac{64}{3}$$ h^2 - 4h + 4 + k^2 - 6k + 9 = frac643$$h^2 - 4h + 4 + k^2 - 6k + 9 = \frac{64}{3}$$ 3(h^2 + k^2 - 4h - 6k + 13) = 64$$3(h^2 + k^2 - 4h - 6k + 13) = 64$$ 3h^2 + 3k^2 - 12h - 18k + 39 - 64 = 0$$3h^2 + 3k^2 - 12h - 18k + 39 - 64 = 0$$ 3(h^2 + k^2) - 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(x^2 + y^2) - 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/sin(alpha/2)$1/\sin(\alpha/2)$ or 1/cos(theta)$1/\cos(\theta)$ depending on whether the angle is measured at intersection or center. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections Class 11 Maths: Straight Lines

### Related Formula textPerpendicular bisector properties: m_1m_2 = -1 text and mid-point lies on the line.$$\text{Perpendicular bisector properties: } m_1m_2 = -1 \text{ and mid-point lies on the line.}$$ textParametric point on circle x^2+y^2=r^2 text is (rcostheta, rsintheta)$$\text{Parametric point on circle } x^2+y^2=r^2 \text{ is } (r\cos\theta, r\sin\theta)$$ ### Core Logic

Let P = (2costheta, 2sintheta)$P = (2\cos\theta, 2\sin\theta)$. Let Q$Q$ on the line 5x + y + 2 = 0$5x + y + 2 = 0$ be Q(alpha, -5alpha-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$-1$. frac2sintheta - (-5alpha - 2)2costheta - alpha = -1$$\frac{2\sin\theta - (-5\alpha - 2)}{2\cos\theta - \alpha} = -1$$ 2sintheta + 5alpha + 2 = -2costheta + alpha$$2\sin\theta + 5\alpha + 2 = -2\cos\theta + \alpha$$ sintheta + costheta + 2alpha + 1 = 0 quad dots (1)$$\sin\theta + \cos\theta + 2\alpha + 1 = 0 \quad \dots (1)$$ ### Step 2: Apply Midpoint Condition Midpoint of PQ$PQ$ is left( frac2costheta + alpha2, frac2sintheta - 5alpha - 22 right)$\left( \frac{2\cos\theta + \alpha}{2}, \frac{2\sin\theta - 5\alpha - 2}{2} \right)$. Substitute into x - y + 1 = 0$x - y + 1 = 0$: frac2costheta + alpha2 - frac2sintheta - 5alpha - 22 + 1 = 0$$\frac{2\cos\theta + \alpha}{2} - \frac{2\sin\theta - 5\alpha - 2}{2} + 1 = 0$$ 2costheta + alpha - 2sintheta + 5alpha + 2 + 2 = 0$$2\cos\theta + \alpha - 2\sin\theta + 5\alpha + 2 + 2 = 0$$ costheta - sintheta + 3alpha + 2 = 0 quad dots (2)$$\cos\theta - \sin\theta + 3\alpha + 2 = 0 \quad \dots (2)$$ ### Step 3: Eliminate alpha and Solve From (1), 2alpha = -sintheta - costheta - 1 implies alpha = frac-sintheta - costheta - 12$2\alpha = -\sin\theta - \cos\theta - 1 \implies \alpha = \frac{-\sin\theta - \cos\theta - 1}{2}$. Substitute alpha$\alpha$ into (2): costheta - sintheta + 3left( frac-sintheta - costheta - 12 right) + 2 = 0$$\cos\theta - \sin\theta + 3\left( \frac{-\sin\theta - \cos\theta - 1}{2} \right) + 2 = 0$$ 2costheta - 2sintheta - 3sintheta - 3costheta - 3 + 4 = 0$$2\cos\theta - 2\sin\theta - 3\sin\theta - 3\cos\theta - 3 + 4 = 0$$ -costheta - 5sintheta + 1 = 0 implies costheta + 5sintheta = 1$$-\cos\theta - 5\sin\theta + 1 = 0 \implies \cos\theta + 5\sin\theta = 1$$ Let's express in half angles: 1 - 2sin^2fractheta2 + 10sinfractheta2cosfractheta2 = 1$$1 - 2\sin^2\frac{\theta}{2} + 10\sin\frac{\theta}{2}\cos\frac{\theta}{2} = 1$$ 2sinfractheta2 left( 5cosfractheta2 - sinfractheta2 right) = 0$$2\sin\frac{\theta}{2} \left( 5\cos\frac{\theta}{2} - \sin\frac{\theta}{2} \right) = 0$$ So, sinfractheta2 = 0 implies costheta = 1$\sin\frac{\theta}{2} = 0 \implies \cos\theta = 1$ or tanfractheta2 = 5 implies costheta = frac1 - tan^2(theta/2)1 + tan^2(theta/2) = frac1 - 251 + 25 = -frac2426 = -frac1213$\tan\frac{\theta}{2} = 5 \implies \cos\theta = \frac{1 - \tan^2(\theta/2)}{1 + \tan^2(\theta/2)} = \frac{1 - 25}{1 + 25} = -\frac{24}{26} = -\frac{12}{13}$. ### Step 4: Final Calculation The abscissa of P$P$ is 2costheta$2\cos\theta$. Values of abscissa are 2(1) = 2$2(1) = 2$ and 2left(-frac1213right) = -frac2413$2\left(-\frac{12}{13}\right) = -\frac{24}{13}$. Sum of abscissa values = 2 - frac2413 = frac26 - 2413 = frac213$= 2 - \frac{24}{13} = \frac{26 - 24}{13} = \frac{2}{13}$. We need 13 times (textSum) = 13 times frac213 = 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 (m_1m_2=-1$m_1m_2=-1$) and midpoint logic simultaneously to create a trigonometric linear equation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Circles Class 11 Maths: Straight Lines

### Related Formula textTwo circles intersect at distinct points if |r_1 - r_2| < d < r_1 + r_2$$\text{Two circles intersect at distinct points if } |r_1 - r_2| < d < r_1 + r_2$$ where d$d$ is the distance between their centers. ### Core Logic Circle 1: (x + 1)^2 + (y + 4)^2 = r^2$(x + 1)^2 + (y + 4)^2 = r^2$ Center C_1 = (-1, -4)$C_1 = (-1, -4)$ and Radius r_1 = r$r_1 = r$. Circle 2: x^2 + y^2 - 4x - 2y - 4 = 0 implies (x - 2)^2 + (y - 1)^2 = 3^2$x^2 + y^2 - 4x - 2y - 4 = 0 \implies (x - 2)^2 + (y - 1)^2 = 3^2$ Center C_2 = (2, 1)$C_2 = (2, 1)$ and Radius r_2 = 3$r_2 = 3$. ### Step 1: Distance Between Centers Distance d$d$ between C_1$C_1$ and C_2$C_2$: d = sqrt(2 - (-1))^2 + (1 - (-4))^2$$d = \sqrt{(2 - (-1))^2 + (1 - (-4))^2}$$ d = sqrt3^2 + 5^2 = sqrt9 + 25 = sqrt34$$d = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$$ ### Step 2: Applying the Intersection Condition For two distinct intersection points: |r - 3| < sqrt34 < r + 3$$|r - 3| < \sqrt{34} < r + 3$$ Breaking this down into two inequalities: 1) |r - 3| < sqrt34 implies -sqrt34 < r - 3 < sqrt34 implies 3 - sqrt34 < r < 3 + sqrt34$|r - 3| < \sqrt{34} \implies -\sqrt{34} < r - 3 < \sqrt{34} \implies 3 - \sqrt{34} < r < 3 + \sqrt{34}$ 2) r + 3 > sqrt34 implies r > sqrt34 - 3$r + 3 > \sqrt{34} \implies r > \sqrt{34} - 3$ Since radius r > 0$r > 0$, taking the intersection of the conditions: r in (sqrt34 - 3, sqrt34 + 3)$$r \in (\sqrt{34} - 3, \sqrt{34} + 3)$$ Thus, alpha = sqrt34 - 3$\alpha = \sqrt{34} - 3$ and beta = sqrt34 + 3$\beta = \sqrt{34} + 3$. ### Step 3: Calculating Final Product alphabeta = (sqrt34 - 3)(sqrt34 + 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: |r_1 - r_2| < d < r_1 + r_2$|r_1 - r_2| < d < r_1 + r_2$. Solving this naturally yields an interval (alpha, beta)$(\alpha, \beta)$ formatted as a difference of squares upon multiplication. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Circles

### Related Formula The equation of a family of curves passing through the intersection of two conics S_1 = 0$S_1 = 0$ and S_2 = 0$S_2 = 0$ is S_1 + lambda S_2 = 0$S_1 + \lambda S_2 = 0$. For this resulting curve to be a circle, the coefficient of x^2$x^2$ must equal the coefficient of y^2$y^2$, and the coefficient of the xy$xy$ term must be zero. ### Core Logic Let the two ellipses be: S_1 equiv x^2 + 2y^2 - 6x - 12y + 23 = 0$S_1 \equiv x^2 + 2y^2 - 6x - 12y + 23 = 0$ S_2 equiv 4x^2 + 2y^2 - 20x - 12y + 35 = 0$S_2 \equiv 4x^2 + 2y^2 - 20x - 12y + 35 = 0$ Equation of the curve passing through their intersection is S_1 + lambda S_2 = 0$S_1 + \lambda S_2 = 0$: (1 + 4lambda)x^2 + (2 + 2lambda)y^2 - (6 + 20lambda)x - (12 + 12lambda)y + (23 + 35lambda) = 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 x^2$x^2$ = coefficient of y^2$y^2$: 1 + 4lambda = 2 + 2lambda implies 2lambda = 1 implies lambda = frac12$$1 + 4\lambda = 2 + 2\lambda \implies 2\lambda = 1 \implies \lambda = \frac{1}{2}$$ ### Step 1: Finding Circle Parameters Substitute lambda = 1/2$\lambda = 1/2$ back into the family equation: (1 + 2)x^2 + (2 + 1)y^2 - (6 + 10)x - (12 + 6)y + left(23 + frac352right) = 0$$(1 + 2)x^2 + (2 + 1)y^2 - (6 + 10)x - (12 + 6)y + \left(23 + \frac{35}{2}\right) = 0$$ 3x^2 + 3y^2 - 16x - 18y + frac812 = 0$$3x^2 + 3y^2 - 16x - 18y + \frac{81}{2} = 0$$ Dividing by 3 to write in standard form: x^2 + y^2 - frac163x - 6y + frac272 = 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 = frac83, quad b = 3$$a = \frac{8}{3}, \quad b = 3$$ The radius r$r$ is given by r^2 = g^2 + f^2 - c$r^2 = g^2 + f^2 - c$: r^2 = left(frac-83right)^2 + (-3)^2 - frac272 = frac649 + 9 - frac272 = frac128 + 162 - 24318 = frac4718$$r^2 = \left(\frac{-8}{3}\right)^2 + (-3)^2 - \frac{27}{2} = \frac{64}{9} + 9 - \frac{27}{2} = \frac{128 + 162 - 243}{18} = \frac{47}{18}$$ ### Step 2: Final Calculation We need to find ab + 18r^2$ab + 18r^2$: ab = left(frac83right)(3) = 8$$ab = \left(\frac{8}{3}\right)(3) = 8$$ 18r^2 = 18left(frac4718right) = 47$$18r^2 = 18\left(\frac{47}{18}\right) = 47$$ ab + 18r^2 = 8 + 47 = 55$$ab + 18r^2 = 8 + 47 = 55$$ ### Pattern Recognition When intersection points of two 2^textnd$2^{\text{nd}}$ degree curves form a circle, apply S_1 + lambda S_2 = 0$S_1 + \lambda S_2 = 0$ immediately, forcing the necessary symmetric coefficients to extract lambda$\lambda$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections Class 11 Maths: Circles

### Core Logic

Equation of circle C_2$C_{2}$ with diameter along y=x$y=x$ passing through origin and having length 10. Wait, C_1$C_1$ has diameter 10. The chord y=x$y=x$ passes through (0,0)$(0,0)$. For the chord to be a diameter of C_2$C_2$, the points of intersection with C_1$C_1$ must form the diameter. The center of C_2$C_2$ 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 sqrt50$\sqrt{50}$? Wait, the problem implies the chord of C_1$C_1$ is y=x$y=x$. If C_1$C_1$ is a circle in x geq 0$x \geq 0$ of diameter 10 through origin. Center of C_2$C_2$ lies on the chord y=x$y=x$. The equation of circle C_2$C_{2}$ is: x^2 + y^2 - 5x - 5y = 0$$x^2 + y^2 - 5x - 5y = 0$$ Its center is Nleft(frac52, frac52right)$N\left(\frac{5}{2}, \frac{5}{2}\right)$. ### Step 1: Find Farthest Chord We need the chord of C_2$C_2$ passing through B(2, 3)$B(2, 3)$ which is farthest from the center Nleft(frac52, frac52right)$N\left(\frac{5}{2}, \frac{5}{2}\right)$. The farthest chord passing through a given point is always perpendicular to the line joining the center to that point. Slope of line NB$NB$: m_NB = frac3 - frac522 - frac52 = fracfrac12-frac12 = -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 m_NB$m_{NB}$: textSlope 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, quad 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Circles Class 11 Mathematics: Straight Lines