If the circles x 2+y 2+5kx+2y+k 0
Web19 mei 2024 · If the circles x2 + y2 + 5Kx + 2y + K = 0 and 2 (x2 + y2 ) + 2Kx + 3y –1 = 0, (K∈R), intersect at the points P and Q, then the line 4x + 5y – K = 0 passes through P and Q, for : (1) exactly two values of K (2) no value of K (3) exactly one value of K (4) infinitely many values of K jee mains 2024 Share It On Facebook Twitter 1 Answer +1 vote Web31 jan. 2024 · Answer is (B) p>0 The parabola y^2=4x is in Q1 and Q2 and passes through (0,0). Further, equation of circle x^2+y^2+2px=0 can be written as (x+p)^2+y^2=p^2 and hence center of circle is (-p,0) and radius is p. Therefore, the circle will touch the parabola externally if centre is p>0 If p<0, circle touches internally as centre is (-p,0) Below we …
If the circles x 2+y 2+5kx+2y+k 0
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WebIf the circles x2+y2+5Kx+2y+K = 0 and 2(x2+y2)+2Kx+3y−1 =0,(K ∈R), intersect at the points P and Q, then the line 4x+5y−K =0 passes through P and Q, for : A exactly one value of K B exactly two values of K C infinitely many values of K D no value of K Solution The correct option is D no value of K Web27 mrt. 2024 · Answer If the circle x 2 + y 2 + 5 K x + 2 y + K = 0 and 2 ( x 2 + y 2) + 2 K x + 3 y − 1 = 0, ( K ∈ R) intersect at the points P and Q, then the line 4 x + 5 y − K = 0 passes through P and Q for: A. exactly two values of K B. exactly one value of K C. no value of K D. infinitely many values of K Last updated date: 27th Mar 2024 •
WebIn all cases a point on the circle follows the rule x 2 + y 2 = radius 2. We can use that idea to find a missing value. Example: x value of 2, and a radius of 5. Start with: x 2 + y 2 = r … WebAnswer (1 of 4): The approach is the same as dealing with a pair of simultaneous equations, where the value of the determinant determines the number of values. If it is 0, such as for x+y=1 and 2x+2y=5, no solutions exist, and if one equation is the same as the other, such as 3x+y=1 and 6x+2y=2,...
WebIf the circles x 2 + y 2 + 5Kx + 2y + K = 0 and 2 (x 2 + y 2) + 2Kx + 3y –1 = 0, (K ∈ R), intersect at the points P and Q, then the line 4x + 5y – K = 0 passes through P and Q, for : A exactly two values of K B no value of K C exactly one value of K D infinitely many values of K Check Answer 2 JEE Main 2024 (Online) 10th April Morning Slot Web3 jan. 2024 · Online Questions and Answers in Analytic Geometry: Points, Lines and Circles Series. Following is the list of practice exam test questions s in this brand new series: MCQ in Analytic Geometry: Points, Lines and Circles. PART 1: MCQ from Number 1 – 50 Answer key: PART 1. PART 2: MCQ from Number 51 – 100 Answer key: PART 2.
Web20 mrt. 2011 · 若关于x,y的二元一次方程组{x+y=5kx-y=9k的解也是二元一次方程2x+3y=6，则k为？ 求 ... 解得x=0，y=0 所以k=0 ... 关注. 展开全部. 2x=14k 2y=-4k ，然后将代入而与一次方程即可，k=3/4
WebNumber of Common Tangents to Two Circles in Different Conditions If the circle... Question If the circles x2+y2−16x−20y+164 =r2 and (x−4)2+(y−7)2 =36 intersect at two distinct … frozen restaurant dish industryWeb If the circles x 2 + y 2 + 5Kx + 2y + K = 0 and 2 (x 2 + y 2) + 2Kx + 3y – 1 = 0, (K ∈ R), intersect at the points P and Q, then the line 4x + 5y – K = 0 passes through P and Q, for: A. Infinitely many values of K B. No value of K C. Exactly two values of K D. Exactly one value of K Please scroll down to see the correct answer and solution guide. frozenrewardsclub.comWebx2 (y-4) 2=16 The blue curves are your hyperbola. The format for the equation of a circle is (x - h)² + (y - k)² = r²; (h. k) is the center. The equation of a circle that I think you … frozen rewardsWeb10th IMP Practice - Read online for free. ... MATHEMATICS IMPORTANT QUESTIONS 2024- 2024 X CLASS. 1 MARK QUESTIONS 1. PAIR OF LINEAR EQUATIONS IN TWO … giantz garden shed 2.38x1.31mWebIf the circles `x^(2)+y^(2) +5Kx+2y + K=0 and2(x^(2)+y^(2))+2Kx +3y-1=0, (KinR)`, intersect at the points P and Q, then the line 4x + 5y - K = 0 passes t... AboutPressCopyrightContact... frozen reunion fandubWeb16 + O (4) = 0 O = Œ4 So (x Œ1) 2 + (y Œ1) Œ4(x Œy) = 0 x2 + y2 Œ 6x + 2y + 2 = 0 r = 22 (correct key is 3) 18. Official Ans. by NTA (4) Sol. Y (h, k) r = h x 22+ y = 1 X h22 k h 1 x2 + y2 = x2 + 1 + 2x y2 = 1 + 2x y = 1 2x; x t 0. 19. Official Ans. by NTA (2) Sol. 5 12 5 12 x x Let length of common chord = 2x 25 x 144 x 1322 after ... frozen retailingWebAnswer (1 of 4): x^2 +y^2 + 6x -6y +5 =0 x^2 +y^2 + 6x -6y = -5 x^2 + 6x +9 +y^2 -6y +9 = 9 + 9 -5 (x+3)^2 +(y-3)^2 =13 So, (-3,3) is centre of circle. Therefore, it should be on … giantz electric winch