If the radical axis of the circles x²+y²+2 g x+2 f y+c=0 and 2 x²+2 y²+3 x+8 y+2 c=0 touches the circle x²+y²+2 x+2 y+1=0, then [AP EAPCET 24 May 2025 S1]

Correct Answer is : either g=34g=3/4g=43 or f=2f=2f=2

Correct Answer is : either g=34g=3/4g=43 or f=2f=2f=2

Test Index

Circle

Section: Mathematics

Question : 1 of 89

Marks: +1, -0

x^2+y^2+2 g x+2 f y+c=0

2 x^2+2 y^2+3 x+8 y+2 c=0

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

Solution:

s_1: x^2+y^2+2 g x+2 f y+c=0

s_2: 2 x^2+2 y^2+3 x+8 y+2 c=0

\Rightarrow \;\; x^2+y^2+\frac{3}{2} x+4 y+c=0

s_3: x^2+y^2+2 x+2 y+1=0

Radical axis

s_1-s_2=0

\Rightarrow (x^2+y^2+2 g x+2 f y+c)

\; - \left(x^2+y^2+\frac{3}{2} x+4 y+c\right)+0

\Rightarrow 2 g x+2 f y-\frac{3 x}{2}-4 y=0

\Rightarrow \left(2 g-\frac{3}{2}\right) x+(2 f-4) y=0

\Rightarrow (4 g-3) x+(4 f-8) y=0

The radical axis touches the circle

s_3

, where radius of

s_3

=\sqrt{(-1)^2+(-1)^2-1}=1 \;\text{ and center is }\;(-1,-1)

The perpendicular distance from the center of

s_3

to the radical axis

=

radius of

s_3

\;\;\frac{\left|(4 g-3) \cdot(-1)+(4 f-8)(-1)\right|}{\sqrt{(4 g-3)^2+(4 f-8)^2}}=1

\;\Rightarrow \;\;\frac{\left|-4 g+3-4 f+8\right|}{\sqrt{(4 g-3)^2+(4 f-8)^2}}=1

\;\Rightarrow \;\;\left|-4 g-4 f+11\right|

\;=\sqrt{(4 g-3)^2+(4 f-8)^2}

\;\Rightarrow \;\;(-4 g-4 f+11)^2

\;=(4 g-3)^2+(4 f-8)^2

\;\Rightarrow \;\;16 g^2+16 f^2+121+32 g f-88 g-88 f

\;=16 g^2-24 g+9+16 f^2-64 f+64

\;\Rightarrow \;\;32 g f-64 g-24 f+48=0

\;\Rightarrow \;\;8 g f-16 g-6 f+12=0

\;\Rightarrow \;\;8 g(f-2)-6(f-2)=0

\;\Rightarrow \;\;(8 g-6)(f-2)=0

\;\Rightarrow \;\;4 g-3)(f-2)=0

\;\Rightarrow \;\;g=\frac{3}{4} \;\text{ or }\; f=2

Either

g=\frac{3}{4}

f=2

is the required condition.

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