Q69.The equation of the circle passing through the point (1, 2) and through the points of intersection of x2 + y2 −4x −6y −21 = 0 and 3x + 4y + 5 = 0 is given by (1) x2 + y2 + 2x + 2y + 11 = 0 (2) x2 + y2 −2x + 2y −7 = 0 (3) x2 + y2 + 2x −2y −3 = 0 (4) x2 + y2 + 2x + 2y −11 = 0
What This Question Tests
This question applies the concept of a family of circles (S + λL = 0) where 'S' is the given circle and 'L' is the given line. The value of λ is determined by using the additional point (1,2) through which the required circle passes.
Concepts Tested
Formulas Used
S + λL = 0
📚 NCERT Sections This Tests
5.11 — Draw All The Isomers (Geometrical And Optical) Of:
Chemistry Class 11 · Chapter 5
5.11 Draw all the isomers (geometrical and optical) of: (i) [CoCl2(en)2] + (ii) [Co(NH3)Cl(en)2] 2+ (iii) [Co(NH3)2Cl2(en)]+
9.17 — (A) Sin I¢C = 1.44/1.68 Which Gives I¢C = 59°. Total Internal Reflection
Physics Class 12 · Chapter 9
9.17 (a) sin i¢c = 1.44/1.68 which gives i¢c = 59°. Total internal reflection takes place when i > 59° or when r < rmax = 31°. Now, (sin i /sin r max max ) = 1.68 , which gives imax ~ 60°. Thus, all incident rays of angles in the range 0 < i < 60° will suffer total internal reflections in the pipe. (If the length of the pipe is finite, which it is in practice, there will be a lower limit on i determined by the ratio of the diameter to the length of the pipe.) (b) If there is no outer coating, i¢c = sin–1(1/1.68) = 36.5°. Now, i = 90° will have r = 36.5° and i¢ = 53.5° which is greater than i¢c. Thus, all incident rays (in the range 53.5° < i < 90°) will suffer total internal reflections.
5.12 — Write All The Geometrical Isomers Of [Pt(Nh3)(Br)(Cl)(Py)] And How Many Of
Chemistry Class 11 · Chapter 5
5.12 Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?
📋 Question Details
- Chapter
- Circles
- Topic
- Family of circles
- Year
- 2012
- Shift
- 07 May Online
- Q Number
- Q69
- Type
- MCQ
- NCERT Ref
- Class 11 Mathematics Ch 11: Conic Sections
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