The circles C : x² + y² + kx + (1+k)y - (k+1) = 0 pass through the same two points ..?

Question

The circles C : x² + y² + kx + (1+k)y - (k+1) = 0 pass through the same two points for every real number k.

(i) find the coordinates of these two points
(ii) find the minimum values of the radius of a circle C

Answer 1

put k = 0

the equation of the resulting circle is

x² + y² + y - 1 = 0



put k = -1

the equation of the resulting circle is

x² + y² - x = 0

x = x² + y²



intersection of these 2 circles will give the required 2 points

x² + y² + y - 1 = 0

x = x² + y²



=>

x + y - 1 = 0

y = 1-x



x = x² + y²

x = x² + 1 + x² -2x

2x² -3x +1 = 0

(2x-1)(x-1) = 0

x = 1 or 1/2

y = 0 or 1/2



the points are (1,0) and (1/2, 1/2)



Radius of a circle of the form

x² + y² + 2gx + 2hy + f = 0

is



√(g² + h² - f)



g = k/2

h = (k+1)/2

f = -(k+1)



radius of the circle is

√[k²/4 + (k+1)²/4 + (k+1)]

= √(k²/2 + k/2 + 1/4 + k + 1)

= √(k²/2 + 3k/2 + 5/4)

= √[(k² + 3k + 5/2)/2]

= √[(k + 3/2)² + 1/4)/2]



for minimum radius put k = -3/2



r min = 1/√8

Answer 2

x² + y² + kx + (1+k)y - (k+1) = 0

Rearrrange to obtain

x² + kx + y²+ (1+k)y = k+1

Next complete the square on the x part and y part to obtain

x² + kx + (k²/4) + y²+ (1+k)y + (1+k)²/4 = k+1 +k²/4 +(1+k)²/4

which gives us standard form

[x + (k/2)]² + [ y+ (1+k)/2]² = k+1 +k²/4 +(1+k)²/4

This is a circle whose center is {-k/2,-(k+1)/2}

and whose radius is k+1 +k²/4 +(1+k)²/4 = {2k²+6k+5}/4

I believe we can find the minimum value for the radius by

expressing y=f(k) = {2k²+6k+5}/4 , determine f'(k) and set

it equal to zero and you should get k= -3/2. substitute this

into f to obtain minimum radius of 1/8

Now to obtain the point all the circles must go through

use two different values for k and solve the system

at k = 1

x² + y² + 1 x + 2y = 2

and at k = -1

x² + y² + -1 x + 0y = 0

subtract these to obtain

2x + 2y = 2

use a different value of k, say k=0 to obtain

x² + y² + 0 x + 1y = 1

subtract equation for k=2 which is

x² + y² + 2 x + 3y = 3

and obtain -2x -2y = -2

both equations are multiples of

x+y=1

I don't have time to move further on this, I hope it gives you some ideas

Answer 3

(a million) to discover an (x,y) for which it is actual regardless of ok's value, we ought to do away with all the ok's from the equation. ok(x + y -a million) + x^2 + y^2 + y -a million = 0 The ok drops out whilst x + y - a million = 0, Leaving x^2 + y^2 + y -a million = 0. the two innovations of that pair of equations are on each and every circle of sort C. replace attitude: put in any 2 values for ok, mutually with 0 and -a million, and discover the intersection of those circles. (2) rewrite the equation of the circle in prevalent sort, (x-a)^2 + (y-b)^2 = r ^2 (x + ok/2)^2 + (y + (a million+ok)/2)^2 = a million/2 ok^2 + 3/2 ok + 5/4 The minimum radius is the squareroot of the minimum of the suited part of that equation.

Answer 4

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let k=0

x^2+y^2+y-1=0

let k=-1

x^2+y^2-x=0

solve and get x= 1/2,1

y=1/2,0

the 2 points are(1/2,1/2)(1,0)



to find radius

use completing the square

x^2+kx+(k/2)^2-(k/2)^2+y^2

+(1+k)y+(1+k/2)^2-(1+k/2)^2-(1+k)=0

remove the constant value and move to the right side

sqrt of[(k/2)^2+(1+k/2)^2+1+k]

is the radius

(k^2+1+k^2+2k+4+4k)/4 must be min

2k^2+5k+6 must be min

take der

4k+5=0

k=-5/4

radius=23/8