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Question: If the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ cuts the three circles $x^2 + y^2 - 5 = 0, x^2 + y^2 -...

If the circle x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0 cuts the three circles x2+y2−5=0,x2+y2−8x−6y+10=0x^2 + y^2 - 5 = 0, x^2 + y^2 - 8x - 6y + 10 = 0 and x2+y2−4x+2y−2=0x^2 + y^2 - 4x + 2y - 2 = 0 at the extremities of their diameters, then

A

C=−5C = -5

B

fg=147/25fg = 147/25

C

g+2f=c+2g + 2f = c + 2

D

4f=3g4f = 3g

Answer

A, B, D

Explanation

Solution

The problem states that a circle S:x2+y2+2gx+2fy+c=0S: x^2 + y^2 + 2gx + 2fy + c = 0 cuts three other circles S1,S2,S3S_1, S_2, S_3 at the extremities of their diameters.

Understanding the condition:

If a circle SS cuts another circle SiS_i at the extremities of its diameter, it means that the common chord of SS and SiS_i is a diameter of SiS_i. This implies that the center of SiS_i must lie on the common chord of SS and SiS_i.

Let the equation of circle SS be x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0.
Let the equation of circle SiS_i be x2+y2+2gix+2fiy+ci=0x^2 + y^2 + 2g_ix + 2f_iy + c_i = 0.
The equation of the common chord of SS and SiS_i is S−Si=0S - S_i = 0, which simplifies to:
(2g−2gi)x+(2f−2fi)y+(c−ci)=0(2g - 2g_i)x + (2f - 2f_i)y + (c - c_i) = 0.
The center of circle SiS_i is Ci(−gi,−fi)C_i(-g_i, -f_i).
For the common chord to be a diameter of SiS_i, its center CiC_i must lie on the common chord. Substituting x=−gix = -g_i and y=−fiy = -f_i into the common chord equation:
(2g−2gi)(−gi)+(2f−2fi)(−fi)+(c−ci)=0(2g - 2g_i)(-g_i) + (2f - 2f_i)(-f_i) + (c - c_i) = 0
−2ggi+2gi2−2ffi+2fi2+c−ci=0-2gg_i + 2g_i^2 - 2ff_i + 2f_i^2 + c - c_i = 0
Rearranging the terms, we get the condition:
2ggi+2ffi−c=2gi2+2fi2−ci2gg_i + 2ff_i - c = 2g_i^2 + 2f_i^2 - c_i.
Since Ri2=gi2+fi2−ciR_i^2 = g_i^2 + f_i^2 - c_i (where RiR_i is the radius of SiS_i), the condition can also be written as:
2ggi+2ffi−c=2Ri2+ci2gg_i + 2ff_i - c = 2R_i^2 + c_i.

Applying the condition to each circle:

  1. For the circle S1:x2+y2−5=0S_1: x^2 + y^2 - 5 = 0
    Here, g1=0,f1=0,c1=−5g_1 = 0, f_1 = 0, c_1 = -5.
    Radius squared R12=g12+f12−c1=02+02−(−5)=5R_1^2 = g_1^2 + f_1^2 - c_1 = 0^2 + 0^2 - (-5) = 5.
    Applying the condition 2gg1+2ff1−c=2R12+c12gg_1 + 2ff_1 - c = 2R_1^2 + c_1:
    2g(0)+2f(0)−c=2(5)+(−5)2g(0) + 2f(0) - c = 2(5) + (-5)
    −c=10−5-c = 10 - 5
    −c=5⇒c=−5-c = 5 \Rightarrow c = -5.

  2. For the circle S2:x2+y2−8x−6y+10=0S_2: x^2 + y^2 - 8x - 6y + 10 = 0
    Here, g2=−4,f2=−3,c2=10g_2 = -4, f_2 = -3, c_2 = 10.
    Radius squared R22=g22+f22−c2=(−4)2+(−3)2−10=16+9−10=15R_2^2 = g_2^2 + f_2^2 - c_2 = (-4)^2 + (-3)^2 - 10 = 16 + 9 - 10 = 15.
    Applying the condition 2gg2+2ff2−c=2R22+c22gg_2 + 2ff_2 - c = 2R_2^2 + c_2:
    2g(−4)+2f(−3)−c=2(15)+102g(-4) + 2f(-3) - c = 2(15) + 10
    −8g−6f−c=30+10-8g - 6f - c = 30 + 10
    −8g−6f−c=40⇒8g+6f+c+40=0-8g - 6f - c = 40 \Rightarrow 8g + 6f + c + 40 = 0.

  3. For the circle S3:x2+y2−4x+2y−2=0S_3: x^2 + y^2 - 4x + 2y - 2 = 0
    Here, g3=−2,f3=1,c3=−2g_3 = -2, f_3 = 1, c_3 = -2.
    Radius squared R32=g32+f32−c3=(−2)2+(1)2−(−2)=4+1+2=7R_3^2 = g_3^2 + f_3^2 - c_3 = (-2)^2 + (1)^2 - (-2) = 4 + 1 + 2 = 7.
    Applying the condition 2gg3+2ff3−c=2R32+c32gg_3 + 2ff_3 - c = 2R_3^2 + c_3:
    2g(−2)+2f(1)−c=2(7)+(−2)2g(-2) + 2f(1) - c = 2(7) + (-2)
    −4g+2f−c=14−2-4g + 2f - c = 14 - 2
    −4g+2f−c=12⇒4g−2f+c+12=0-4g + 2f - c = 12 \Rightarrow 4g - 2f + c + 12 = 0.

Solving the system of equations:

We have the following system of equations:

  1. c=−5c = -5
  2. 8g+6f+c+40=08g + 6f + c + 40 = 0
  3. 4g−2f+c+12=04g - 2f + c + 12 = 0

Substitute c=−5c = -5 into equations (2) and (3): 2. 8g+6f−5+40=0⇒8g+6f+35=08g + 6f - 5 + 40 = 0 \Rightarrow 8g + 6f + 35 = 0 3. 4g−2f−5+12=0⇒4g−2f+7=04g - 2f - 5 + 12 = 0 \Rightarrow 4g - 2f + 7 = 0

From equation (3), multiply by 3:
3(4g−2f+7)=0⇒12g−6f+21=03(4g - 2f + 7) = 0 \Rightarrow 12g - 6f + 21 = 0.
Add this to equation (2):
(8g+6f+35)+(12g−6f+21)=0(8g + 6f + 35) + (12g - 6f + 21) = 0
20g+56=020g + 56 = 0
20g=−56⇒g=−5620=−14520g = -56 \Rightarrow g = -\frac{56}{20} = -\frac{14}{5}.

Substitute g=−145g = -\frac{14}{5} into 4g−2f+7=04g - 2f + 7 = 0:
4(−145)−2f+7=04(-\frac{14}{5}) - 2f + 7 = 0
−565−2f+7=0-\frac{56}{5} - 2f + 7 = 0
−2f=565−7-2f = \frac{56}{5} - 7
−2f=56−355-2f = \frac{56 - 35}{5}
−2f=215⇒f=−2110-2f = \frac{21}{5} \Rightarrow f = -\frac{21}{10}.

So, the values are c=−5c = -5, g=−14/5g = -14/5, f=−21/10f = -21/10.

Checking the options:

A. C=−5C = -5: This matches our calculated value of cc. So, A is correct.

B. fg=147/25fg = 147/25:
fg=(−145)×(−2110)=14×215×10=29450=14725fg = (-\frac{14}{5}) \times (-\frac{21}{10}) = \frac{14 \times 21}{5 \times 10} = \frac{294}{50} = \frac{147}{25}. So, B is correct.

C. g+2f=c+2g + 2f = c + 2:
LHS: g+2f=−145+2(−2110)=−145−4210=−145−215=−355=−7g + 2f = -\frac{14}{5} + 2(-\frac{21}{10}) = -\frac{14}{5} - \frac{42}{10} = -\frac{14}{5} - \frac{21}{5} = -\frac{35}{5} = -7.
RHS: c+2=−5+2=−3c + 2 = -5 + 2 = -3.
Since −7≠−3-7 \neq -3, option C is incorrect.

D. 4f=3g4f = 3g:
LHS: 4f=4(−2110)=−8410=−4254f = 4(-\frac{21}{10}) = -\frac{84}{10} = -\frac{42}{5}.
RHS: 3g=3(−145)=−4253g = 3(-\frac{14}{5}) = -\frac{42}{5}.
Since LHS = RHS, option D is correct.

Conclusion:
Options A, B, and D are correct.