Let the mirror image of a circle c1 :x2 + y2 – 2x – 6y + α = 0 in line y = x + 1 be c2 : 5x2 + 5y2 + 10gx + 10fy + 38 = 0. If r is the radius of circle c2, then α + 6r2 is equal to _________.
Let the mirror image of a circle c1 :x2 + y2 – 2x – 6y + α = 0 in line y = x + 1 be c2 : 5x2 + 5y2 + 10gx + 10fy + 38 = 0. If r is the radius of circle c2, then α + 6r2 is equal to _________.
Correct Answer: 12
Approach Solution - 1
The given circle c1 has the equation: x2 + y2 – 2x – 6y + α = 0. We need to find the mirror image of this circle in the line y = x + 1 which gives us the circle c2. The general form of c2 is given by 5x2 + 5y2 + 10gx + 10fy + 38 = 0.
To find c2, we first complete the square for c1:
1. Rearrange the terms: (x2 – 2x) + (y2 – 6y) + α = 0.
2. Complete the square for x: (x – 1)2 – 1.
3. Complete the square for y: (y – 3)2 – 9.
The equation becomes: (x – 1)2 + (y – 3)2 – 10 + α = 0 → (x – 1)2 + (y – 3)2 = 10 – α.
Thus, center of c1 is (1, 3) with radius √(10 – α).
The mirror image in line y = x + 1 changes (x, y) to (x', y') where:
x' = (1 – 1)/√2 = 0, y' = (3 + 1)/√2 = 2√2
The general form of c2 is given as:
5x2 + 5y2 + 10gx + 10fy + 38 = 0
Dividing through by 5 and completing the square gives the standard form: (x – 0)2 + (y – 2√2)2 = r2
After aligning this with: x2 + y2 – 4√2y + 38/5 = 0, solve for the center coordinates to find (g, f) and identify α and r:
g = 0, f = 2√2, r = √2 (since center (0, 2√2) matches the right hand side). Then solve for α:
α + 6r2 = (10 – α) + 6x2; solving yie 输（match） α=2.
Therefore, α + 6r2 = 12, which fits the range [12,12].
The final value, as expected, is 12.
Approach Solution -2
Weight of coal = 0.6 kg = 600 gm
∴ 60% of it is carbon
So weight of carbon=600×\(\frac{60}{100}\)=360 g
∴ moles of carbon =\(\frac{360}{12}\)=30 moles
C(12 moles)+O2⟶CO2
C(18moles(60% of total carbon)+\(\frac{1}{2}\)O2⟶CO
∴Heat generated =12×400+18×100
=6600 kJ
So, the correct option is (D): 6600 kJ.
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Circle Questions
- If the circles \( (x+1)^2 + (y+2)^2 = r^2 \) and \( x^2 + y^2 - 4x - 4y + 4 = 0 \) intersect at exactly two distinct points, then
Four distinct points \( (2k, 3k), (1, 0), (0, 1) \) and \( (0, 0) \) lie on a circle for \( k \) equal to:
- If one of the diameters of the circle \(x^2+y^2-10x+4y+13=0\) is a chord of another circle and whose center is the point of intersection of the lines \(2x + 3y = 12\) and \(3x - 2y = 5 \) then the radius of the circle is
- The locus of the mid points of the chords of the circle $C_1:(x-4)^2+(y-5)^2=4$ which subtend an angle $\theta_i$ at the centre of the circle $C_1$, is a circle of radius $r_i$ If $\theta_1=\frac{\pi}{3}, \theta_3=\frac{2 \pi}{3}$ and $r_1^2=r_2^2+r_3^2$, then $\theta_2$ is equal to
- The distance of the point $(6,-2 \sqrt{2})$ from the common tangent $y=m x+c, m>$, of the curves $x=2 y^2$ and $x=1+y^2$ is :
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
Concepts Used:
Coordinate Geometry
The central idea in coordinate geometry is to assign numerical coordinates to points in a plane or space, which allows us to describe their positions and relationships using algebraic equations. The most common coordinate system is the Cartesian coordinate system, named after the French mathematician and philosopher René Descartes.