### Related Formula For two circles with centers C_1, C_2$C_1, C_2$ and radii r_1, r_2$r_1, r_2$ to intersect at two distinct real points, the distance between their centers must satisfy: |r_1 - r_2| < C_1C_2 < r_1 + r_2$|r_1 - r_2| < C_1C_2 < r_1 + r_2$ ### Core Logic Extracting properties from the circle equations: - Circle C$C$: x^2 + y^2 = 4 implies$x^2 + y^2 = 4 \implies$ Center C_1 = (0,0)$C_1 = (0,0)$, Radius r_1 = 2$r_1 = 2$ - Circle C'$C'$: x^2 + y^2 - 4lambda x + 9 = 0 implies$x^2 + y^2 - 4\lambda x + 9 = 0 \implies$ Center C_2 = (2lambda, 0)$C_2 = (2\lambda, 0)$, Radius r_2 = sqrt(-2lambda)^2 - 9 = sqrt4lambda^2 - 9$r_2 = \sqrt{(-2\lambda)^2 - 9} = \sqrt{4\lambda^2 - 9}$ For real intersection conditions, the radius must be defined: 4lambda^2 - 9 > 0 implies lambda^2 > frac94 quad implies (1)$4\lambda^2 - 9 > 0 \implies \lambda^2 > \frac{9}{4} \quad \implies (1)$ Distance between centers: C_1C_2 = sqrt(2lambda - 0)^2 + 0^2 = |2lambda|$C_1C_2 = \sqrt{(2\lambda - 0)^2 + 0^2} = |2\lambda|$. ### Step 1: Formulate the Triangle Inequality Solutions Applying the intersection condition: |2 - sqrt4lambda^2 - 9| < |2lambda| < 2 + sqrt4lambda^2 - 9$|2 - \sqrt{4\lambda^2 - 9}| < |2\lambda| < 2 + \sqrt{4\lambda^2 - 9}$ The right side inequality |2lambda| < 2 + sqrt4lambda^2 - 9$|2\lambda| < 2 + \sqrt{4\lambda^2 - 9}$ is always valid for real radii. Solving the left side inequality by squaring: left(2 - sqrt4lambda^2 - 9right)^2 < (2lambda)^2$\left(2 - \sqrt{4\lambda^2 - 9}\right)^2 < (2\lambda)^2$ 4 + (4lambda^2 - 9) - 4sqrt4lambda^2 - 9 < 4lambda^2$4 + (4\lambda^2 - 9) - 4\sqrt{4\lambda^2 - 9} < 4\lambda^2$ -5 - 4sqrt4lambda^2 - 9 < 0 implies 4sqrt4lambda^2 - 9 > -5$-5 - 4\sqrt{4\lambda^2 - 9} < 0 \implies 4\sqrt{4\lambda^2 - 9} > -5$ Since a square root is always non-negative, square both sides to get the strict bound: 16(4lambda^2 - 9) > 25 implies 64lambda^2 - 144 > 25$16(4\lambda^2 - 9) > 25 \implies 64\lambda^2 - 144 > 25$ 64lambda^2 > 169 implies lambda^2 > frac16964 quad implies (2)$64\lambda^2 > 169 \implies \lambda^2 > \frac{169}{64} \quad \implies (2)$ Combining bounds (1) and (2), condition (2) is more restrictive, meaning: lambda in left(-infty, -frac138right) cup left(frac138, inftyright)$\lambda \in \left(-\infty, -\frac{13}{8}\right) \cup \left(\frac{13}{8}, \infty\right)$ This can be written as mathbbR - left[-frac138, frac138right]$\mathbb{R} - \left[-\frac{13}{8}, \frac{13}{8}\right]$. ### Step 2: Coordinate Analysis of the Target Point Comparing our interval with mathbbR - [a, b]$\mathbb{R} - [a, b]$, we identify: a = -frac138, quad b = frac138$a = -\frac{13}{8}, \quad b = \frac{13}{8}$ Now, substitute these bounds to locate the coordinates of our target point (8a+12, \, 16b-20)$(8a+12, \, 16b-20)$: - x$x$-coordinate: 8left(-frac138right) + 12 = -13 + 12 = -1$8\left(-\frac{13}{8}\right) + 12 = -13 + 12 = -1$ - y$y$-coordinate: 16left(frac138right) - 20 = 26 - 20 = 6$16\left(\frac{13}{8}\right) - 20 = 26 - 20 = 6$ Hence, the point is (-1, 6)$(-1, 6)$. ### Step 3: Test Point against the Options Substitute (-1, 6)$(-1, 6)$ into the given curves to find a match: Testing Option (4): 6x^2 + y^2 = 42$6x^2 + y^2 = 42$ 6(-1)^2 + (6)^2 = 6(1) + 36 = 42$6(-1)^2 + (6)^2 = 6(1) + 36 = 42$ Since LHS = RHS, the point satisfies the curve in Option (4). ### Pattern Recognition Sees: Dynamic variable interval limits forming loci requirements. Shortcut: When checking inequalities like |2 - sqrtz| < 2lambda$|2 - \sqrt{z}| < 2\lambda$, algebraic squaring helps simplify the terms quickly by eliminating the parameter 4lambda^2$4\lambda^2$ from both sides. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections (Circles) Class 11 Mathematics: Linear Inequalities

### Related Formula - Distance from a point (x_0, y_0)$(x_0, y_0)$ to a straight line Ax + By + C = 0$Ax + By + C = 0$ is: d = frac|Ax_0 + By_0 + C|sqrtA^2 + B^2$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$ - Tangency Condition: Perpendicular distance from the center of a circle to a line must equal its radius (d = r$d = r$). ### Core Logic First, find the point of intersection P$P$ between the circle x^2 + y^2 = 3$x^2 + y^2 = 3$ and the parabola x^2 = 2y$x^2 = 2y$: 2y + y^2 = 3 implies y^2 + 2y - 3 = 0$2y + y^2 = 3 \implies y^2 + 2y - 3 = 0$ (y + 3)(y - 1) = 0 implies y = 1 quad text(since P text lies in the first quadrant)$(y + 3)(y - 1) = 0 \implies y = 1 \quad \text{(since } P \text{ lies in the first quadrant)}$ Substituting y = 1$y = 1$ back: x^2 = 2(1) implies x = sqrt2$x^2 = 2(1) \implies x = \sqrt{2}$. Thus, point P = (sqrt2, 1)$P = (\sqrt{2}, 1)$. ### Step 1: Solve for Line Parameter alpha Since P(sqrt2, 1)$P(\sqrt{2}, 1)$ lies on line L: sqrt2x + y = alpha$L: \sqrt{2}x + y = \alpha$: sqrt2(sqrt2) + 1 = alpha implies 2 + 1 = alpha implies alpha = 3$\sqrt{2}(\sqrt{2}) + 1 = \alpha \implies 2 + 1 = \alpha \implies \alpha = 3$ So the equation of line L$L$ is sqrt2x + y - 3 = 0$\sqrt{2}x + y - 3 = 0$. ### Step 2: Find Centers Q1 and Q2 The centers lie on the y-axis, so let their coordinates be (0, k)$(0, k)$. The perpendicular distance to line L$L$ is equal to the radius 2sqrt3$2\sqrt{3}$: frac|sqrt2(0) + k - 3|sqrt(sqrt2)^2 + 1^2 = 2sqrt3$\frac{|\sqrt{2}(0) + k - 3|}{\sqrt{(\sqrt{2})^2 + 1^2}} = 2\sqrt{3}$ frac|k - 3|sqrt3 = 2sqrt3 implies |k - 3| = 6$\frac{|k - 3|}{\sqrt{3}} = 2\sqrt{3} \implies |k - 3| = 6$ Unfolding the absolute parameter: - k - 3 = 6 implies k = 9 implies Q_1 = (0, 9)$k - 3 = 6 \implies k = 9 \implies Q_1 = (0, 9)$ - k - 3 = -6 implies k = -3 implies Q_2 = (0, -3)$k - 3 = -6 \implies k = -3 \implies Q_2 = (0, -3)$ ### Step 3: Evaluate Triangle Area Squared Using the matrix determinant method for the area of triangle PQ_1Q_2$\triangle PQ_1Q_2$: textArea = frac12 beginvmatrix sqrt2 & 1 & 1 \\ 0 & 9 & 1 \\ 0 & -3 & 1 endvmatrix = frac12 left| sqrt2 cdot (9 - (-3)) right| = 6sqrt2$\text{Area} = \frac{1}{2} \begin{vmatrix} \sqrt{2} & 1 & 1 \\ 0 & 9 & 1 \\ 0 & -3 & 1 \end{vmatrix} = \frac{1}{2} \left| \sqrt{2} \cdot (9 - (-3)) \right| = 6\sqrt{2}$ Squaring the final area value: textArea^2 = (6sqrt2)^2 = 72$\text{Area}^2 = (6\sqrt{2})^2 = 72$ ### Pattern Recognition Sees: Intersecting conics interacting via tangent logic matrix calculations. Shortcut: Since the base Q_1Q_2$Q_1Q_2$ sits entirely on the y-axis, its length is simply 9 - (-3) = 12$9 - (-3) = 12$. The height corresponds to the x-coordinate of point P$P$, which is sqrt2$\sqrt{2}$. Area is frac12 times 12 times sqrt2 = 6sqrt2$\frac{1}{2} \times 12 \times \sqrt{2} = 6\sqrt{2}$, saving determinant steps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections (Circles and Parabolas) Class 10 Coordinate Geometry: Area of Triangles

### Related Formula y = mx - 2am - am^3 quad text(Equation of normal to y^2 = 4ax)$y = mx - 2am - am^3 \quad \text{(Equation of normal to } y^2 = 4ax)$ ### Core Logic First, extract the centre C$C$ of the circle x^2 + y^2 - 4x - 16y + 64 = 0$x^2 + y^2 - 4x - 16y + 64 = 0$: C equiv (2, 8)$C \equiv (2, 8)$ For the parabola y^2 = 4x$y^2 = 4x$, comparing with y^2 = 4ax$y^2 = 4ax$, we get a = 1$a = 1$. The shortest distance between a point and a curve is measured along the normal to the curve passing through that point. We need the normal to the parabola that passes through the circle's centre (2, 8)$(2, 8)$. ### Step 1: Applying Normal Condition Equation of normal in slope form (a=1$a=1$): y = mx - 2m - m^3$y = mx - 2m - m^3$ Since it passes through (2, 8)$(2, 8)$: 8 = m(2) - 2m - m^3$8 = m(2) - 2m - m^3$ 8 = -m^3 Rightarrow m^3 = -8 Rightarrow m = -2$8 = -m^3 \Rightarrow m^3 = -8 \Rightarrow m = -2$ ### Step 2: Finding Coordinate Point on Parabola The foot of the normal on the parabola is given by the coordinate P(am^2, -2am)$P(am^2, -2am)$. Substitute a=1, m=-2$a=1, m=-2$: P equiv (1(-2)^2, -2(1)(-2))$P \equiv (1(-2)^2, -2(1)(-2))$ P equiv (4, 4)$P \equiv (4, 4)$ ### Step 3: Calculating Shortest Distance The distance d$d$ is between the point P(4, 4)$P(4, 4)$ and the centre C(2, 8)$C(2, 8)$: d = sqrt(4-2)^2 + (4-8)^2$d = \sqrt{(4-2)^2 + (4-8)^2}$ d = sqrt2^2 + (-4)^2 = sqrt4 + 16 = sqrt20$d = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20}$ Thus, d^2 = 20$d^2 = 20$. ### Pattern Recognition Shortest distance between a curve and a fixed point (like a circle's center) ALWAYS lies exactly along their common normal. Write the parameterized normal equation and force it through the coordinate to extract the slope root. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Circles Class 11 Maths: Parabola

### Related Formula T = S_1$T = S_1$ Equation of chord with a given midpoint (x_1, y_1)$(x_1, y_1)$ for an ellipse fracx^2a^2 + fracy^2b^2 = 1$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is: fracx x_1a^2 + fracy y_1b^2 = fracx_1^2a^2 + fracy_1^2b^2$\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2}$ ### Core Logic Given the ellipse fracx^225 + fracy^216 = 1$\frac{x^2}{25} + \frac{y^2}{16} = 1$ and midpoint (x_1, y_1) = (1, frac25)$(x_1, y_1) = (1, \frac{2}{5})$. Substitute these into the T = S_1$T = S_1$ formula: fracx(1)25 + fracy(2/5)16 = frac1^225 + frac(2/5)^216$\frac{x(1)}{25} + \frac{y(2/5)}{16} = \frac{1^2}{25} + \frac{(2/5)^2}{16}$ fracx25 + fracy40 = frac125 + frac4/2516$\frac{x}{25} + \frac{y}{40} = \frac{1}{25} + \frac{4/25}{16}$ frac8x + 5y200 = frac125 + frac1100$\frac{8x + 5y}{200} = \frac{1}{25} + \frac{1}{100}$ frac8x + 5y200 = frac5100 = frac10200$\frac{8x + 5y}{200} = \frac{5}{100} = \frac{10}{200}$ This simplifies to the chord equation: 8x + 5y = 10 Rightarrow y = frac10 - 8x5$8x + 5y = 10 \Rightarrow y = \frac{10 - 8x}{5}$ ### Step 1: Finding Points of Intersection Substitute the chord equation back into the ellipse equation to find intersection coordinates: fracx^225 + frac(frac10 - 8x5)^216 = 1$\frac{x^2}{25} + \frac{(\frac{10 - 8x}{5})^2}{16} = 1$ fracx^225 + frac(10 - 8x)^2400 = 1$\frac{x^2}{25} + \frac{(10 - 8x)^2}{400} = 1$ Multiply the entire equation by 400: 16x^2 + (10 - 8x)^2 = 400$16x^2 + (10 - 8x)^2 = 400$ 16x^2 + 100 - 160x + 64x^2 = 400$16x^2 + 100 - 160x + 64x^2 = 400$ 80x^2 - 160x - 300 = 0$80x^2 - 160x - 300 = 0$ Divide by 20: 4x^2 - 8x - 15 = 0$4x^2 - 8x - 15 = 0$ ### Step 2: Solving the Quadratic Solve for roots x_1$x_1$ and x_2$x_2$: x = frac-(-8) pm sqrt(-8)^2 - 4(4)(-15)2(4)$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(4)(-15)}}{2(4)}$ x = frac8 pm sqrt64 + 2408 = frac8 pm sqrt3048$x = \frac{8 \pm \sqrt{64 + 240}}{8} = \frac{8 \pm \sqrt{304}}{8}$ Difference of x-coordinates: |x_1 - x_2| = fracsqrt3044$|x_1 - x_2| = \frac{\sqrt{304}}{4}$ ### Step 3: Calculating Chord Length Because the points lie on the line y = frac10 - 8x5$y = \frac{10 - 8x}{5}$, the difference in y-coordinates relates to the slope m = -frac85$m = -\frac{8}{5}$: |y_1 - y_2| = |-frac85| |x_1 - x_2|$|y_1 - y_2| = |-\frac{8}{5}| |x_1 - x_2|$ The distance D$D$ between the points is: D = sqrt(x_1 - x_2)^2 + (y_1 - y_2)^2 = |x_1 - x_2| sqrt1 + m^2$D = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} = |x_1 - x_2| \sqrt{1 + m^2}$ D = fracsqrt3044 sqrt1 + left(frac-85right)^2$D = \frac{\sqrt{304}}{4} \sqrt{1 + \left(\frac{-8}{5}\right)^2}$ D = fracsqrt3044 sqrt1 + frac6425 = fracsqrt3044 fracsqrt895$D = \frac{\sqrt{304}}{4} \sqrt{1 + \frac{64}{25}} = \frac{\sqrt{304}}{4} \frac{\sqrt{89}}{5}$ D = fracsqrt16 times 194 fracsqrt895 = frac4sqrt194 fracsqrt895 = fracsqrt16915$D = \frac{\sqrt{16 \times 19}}{4} \frac{\sqrt{89}}{5} = \frac{4\sqrt{19}}{4} \frac{\sqrt{89}}{5} = \frac{\sqrt{1691}}{5}$ ### Pattern Recognition For intersecting lines and conics, you rarely need to explicitly find y_1$y_1$ and y_2$y_2$. Use the slope property: Distance = |x_1 - x_2| sqrt1+m^2$= |x_1 - x_2| \sqrt{1+m^2}$ to bypass entirely the substitution of messy quadratic roots back into the linear equation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Ellipse Class 11 Maths: Straight Lines