Question

The radical axis of the circles x²+y²+2gx+2fy+c=0 and 2x²+2y²+3x+8y+2c=0 touches the circle x²+y²+2x-2y+1=0. SHow that either g=3/4 or f=2. 

Answer 1

Radical axis of two circle is defined as locus of a point such that length of tangent from it will be equal .
Let Point (P) : ( a, b)
S1 (a, b) = S2(a , b)
a² + b² + 2ga + 2fb + C = a² + b² +3/2a + 4b + C = 0 [ 2x² + 2y² + 3x + 8y + 2C = 0 divided by 2 for forming circle equation]

a(2g - 3/2) + b(2f - 4)b = 0 this is locus of radical axis.
so, x(2g - 3/2) + y(2f - 4) = 0 is redical axis of given circles.
this touched the x² + y² + 2x -2y + 1 = 0 so, radius =√(1² + 1² -1) and centre =(-1,1)
so,
radius of circle = distance between centre and touching point.
√(1² + 1² - 1) = |(3/2 - 2g)+(2f -4)|/√{(3/2 - 2g)² + (2f -4)²}
take square both sides,

2(3/2 - 2g)(2f -4) = 0
so, (3/2 - 2g) = 0 or (2f -4) = 0
g = 3/4 or f = 2
hence , proved//

Answer 2

The given equation of circle C1 is;x2+y2−2x−2y−7=0and equation of circle C2 isx2+y2+4x+2y+k=0Comparing both equations with standard form i.e. x2+y2+2gx+2fy+c=0, we have;Center of circle C1 is (1,1) and center of circle C2 is (−2,−1)Also radius of circle C1 is R1=12+12−(−7)‾‾‾‾‾‾‾‾‾‾‾‾‾‾√ = 3And radius of circle C2 is R2=(2)2+(1)2−k‾‾‾‾‾‾‾‾‾‾‾‾‾‾√=5−k‾‾‾‾‾√Now since both circles cuts orthogonally so we have;R12+R22=d2 {where d is distance between both centers}⇒32+5−k=[(−2−1)2+(−1−1)2]⇒14−k=13⇒k=1so, R2=5−1‾‾‾‾‾√=2Now AB is common chord of circle.In △APC' we have;AC'2=AP2+PC'2⇒R12=h2+x2 {assuming PC'=x}⇒32=h2+x2⇒h2=9−x2 ...(i)Similarly in △APC;h2=4−(13‾‾‾√−x)2 ...(ii) {Since CC'=d⇒PC=d−x and d=13‾‾‾√}so from (i) and (ii) we have;4−(13‾‾‾√−x)2=9−x2⇒4−(13+x2−213‾‾‾√x)=9−x2⇒4−13−x2+213‾‾‾√x=9−x2⇒213‾‾‾√x=18⇒x=913√then h2=9−(913√)2=3613⇒h=613√Therefore length of common chord AB=2h=2×613√=1213√

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