A variable straight line through A(−1,−1) is drawn to cut the circle x2+y2=1 at the points B,C. If P is chosen on the line ABC such that AB,AP,AC are in A.P then the locus of P is
A
x2+y2+x+y=0
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B
x2+y2−x−y=0
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C
x2+y2+x−y=0
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D
x2+y2−x+y=0
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Solution
The correct option is Ax2+y2+x+y=0 Let locus at point P=(h,k),AP=r, inclination of line ABC be θ then any point on the line will be of form (x,y)=(−1+rcosθ,−1+rsinθ)
For this point to lie on the circle x2+y2=1 ⇒(−1+rcosθ)2+(−1+rsinθ)2=1 ⇒r2−2(cosθ+sinθ)r+1=0 r1,r2 are its roots
Let r1=AB,r2=AC
If AB,AP,AC are in A.P ⇒2AP=(AB+AC) ⇒2r=(r1+r2) ⇒2r=2(cosθ+sinθ) ⇒r2=rcosθ+rsinθ ⇒(h+1)2+(k+1)2=(h+1)+(k+1) ⇒h2+k2+h+k=0
so, locus is x2+y2+x+y=0