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# Q1. Show that for all values of p, the circle x2 + y2 - x(3p+4) –y(p-2) + 10p =0 passes through the point (3,1). If p varies, find the locus of the centre of the above circle.Q2. Whatever be the values of θ, prove that the locus of the point of intersection of the straight lines y =atanθ and asin3θ + ycos3θ = asinθcosθ  is a circle. Find the equation of the circle.Q3. Prove that the square of the distance between the two points (x1 , y1) and    (x2 , y2) of the circle x2 +y2 = a2 is 2(a2-x1x2-y1y2).

## 1 Answers

9 years ago

Q1:

put x=3, y=1 in the eq. U get  9+1-3(3p+4)-(p-2)+10p=12-12-10p+10p=0.

ie. For any value of p the pt.(3,1) satisfy the eq. Hence proved.

Let center be (h,k). From eq.

h=(3p+4)/2   =>  p=(2h-4)/3

k=(p-2)/2 . Sub. value of p we get:

3k-h+5=0

ie. locus of center is:

3y-x+5=0 ie. straight line

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