Objective question in Circle

IIT JEE 2015 Questions › Co-ordinate Geometry questions › Objective question in Circle

The locus of the centre of circle which cuts the circles {{x}^{2}}+{{y}^{2}}+4x-6y+9=0 and {{x}^{2}}+{{y}^{2}}-4x+6y+4=0 orthogonally is

#Mathematics #Co-ordinate Geometry #Circle

UP 0 DOWN 0 0 2
Post on FacebookPost anonymously?

2 Answers

Shahrukh Uddin ·2012-09-30 23:25:59

let the centre be (h,k)
h=-g k=-f
g=-h f=-k ------(1)
now if it intersects orthogonally then from theformula,4g-6f=c=9 and -4g+6f=c+4 then eliminating c we get
8g-12f=5
substituting (1),we get
12k-8h=5
=>8h-12k+5=0
now h=x and k=y
8x-12y+5=0 is the locus of the centre of the circle.but answer went wrong in chakravyuh test please tell where i went wrong.

UP 0 DOWN 0
Post on FacebookPost anonymously?

Shahrukh Uddin ·2012-09-30 23:27:22

answer given there is 12x+8y+5=0

UP 0 DOWN 0
Post on FacebookPost anonymously?

Your Answer

Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

Close [X]

Objective question in Circle

2 Answers

Your Answer

Add Image

Similar Questions