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Question 31088: TI found this problem really tough... ^^U
The circles C: x^2 + y^2 + kx + (1+k)y - (k+1) = 0, pass through the same two points for every real number k
(1) Find the coordinates of these two points
(2) Find the Minimum value of the radius of a circle C
Any help would be really really appreciated!
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! The circles C: x^2 + y^2 + kx + (1+k)y - (k+1) = 0, pass through the same two points for every real number k
(1) Find the coordinates of these two points
LET US CONSIDER TO CIRCLES OF THIS FAMILY FOR K=K1 AND K=K2..THEIR EQNS ARE
x^2 + y^2 + k1x + (1+k1)y - (k1+1) = 0.......................I
x^2 + y^2 + k2x + (1+k2)y -(k2+1) = 0....................II
EQN.I - EQN.II GIVES
X(K1-K2)+y(1+K1-1-K2)-K1-1+K2+1=0
X(K1-K2)+Y(K1-K2)=K1-K2...SINCE K1 AND K2 ARE DIFFERENT K1-K2 IS NOT ZERO.HENCE DIVIDING BY K1-K2,WE GET
X+Y=1..THAT IS THIS LOCUR CALLED RADICAL AXIS IS INDEPENDENT OF K....
SO ALL THE CIRCLES PASS THROUGH 2 FIXED POINTS WHERE THIS RADICAL AXIS CUTS THE ABOVE CIRCLES.
SUBSTITUTING THIS IN EQN OF CIRCLE WE GET
X^2+(1-X)^2+KX+(1+K)(1-X)-(K+1)=0
X^2+1+X^2-2X+KX+1-X+K-KX-K-1=0
2X^2-3X+1=0
(2X-1)(X-1)=0
2X=1...OR...X=1/2...SO Y=1-X=1/2
X=1....SO Y=1-X=0
SO THESE CIRCLES ALWAYS PASS THROUGH (0.5,0.5) AND (1,0) FOR ANY VALUE OF K.
(2) Find the Minimum value of the radius of a circle C
x^2 + y^2 + kx + (1+k)y - (k+1) = 0
={x^2 +2(X)(K/2)+(K/2)^2}+{ y^2 +2(Y)(1+k)/2+(1+K)^2/4}-(K/2)^2-(1+K)^2/4 - (k+1) = 0
{X+K/2}^2+{Y+(1+K)/2}^2=(1/4)*{K^2+(1+K)^2+4(K+1)}=(1/4){K^2+1+K^2+2K+4K+4}
=(1/4){2K^2+6K+5}=(K^2+3K+2.5)/2={(K^2+2K*1.5+1.5^2)-(1.5)^2+2.5}/2
={(K+1.5)^2+1/4}/2
IF THIS TO REPRESENT A CIRCLE THE R.H.S BEING R^2,HAS TO BE POSITIVE,WHICH IS THE CASE AS BOTH TERMS ON RHS ARE POSITIVE.ITS MINIMUM VALUE IS 1/8 WHEN K=-1.5.
HENCE MINIMUM RADIUS OF THE CIRCLE IS SQRT.(1/8)=0.5*SQRT.(0.5)
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