Question

write the condition for the circle bisecting the circumference of another circle​​

Answer 1

Answer:

Let us consider the equation of bisecting circle to be "S1"...and the equation of bisected circle be "S2"

S1: x²+y² +4x+22y+l

S2: x²+y²-2x+8y-m

In order to calculate the value of "l+m"......we first need to calculate the value of common tangent...... Let us denote the eqn of common tangent with "L"

The eqn of common tangent will be = S1-S2

So eqn of common tangent is= x²+y²+4x+22y+l - (x²+y²-2x+8y-m)

Eqn of common tangent = 6x+14y+(l+m)=0

Also the centre of the bisected circle should lie on the common tangent.....

Centre of bisected circle = (1,-4). (-g,-f)

Putting the coordinates of centre of the circle in eqn of common tangent.....

6(1)+14(-4)+(l+m)=0

6-56+(l+m)=0

-50 +(l+m)=0

l+m=50

Explanation:

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Answer 2

Answer:

Hello, Here below is the answer to your question:-

So now:-

S1: x²+y² +4x+22y+l

S2: x²+y²-2x+8y-m

The equation of common tangent will be = S1-S2

So equation of common tangent is= x²+y²+4x+22y+l - (x²+y²-2x+8y-m)

Equation of common tangent = 6x+14y+(l+m)=0

Center of bisected circle = (1,-4). (-g,-f)

6(1)+14(-4)+(l+m)=0

6-56+(l+m)=0

-50 +(l+m)=0

l+m=50

 

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