Question

9.
Suppose f(x, y) = 0 is the equation of the circle such that f(x, 1) = 0 has equal roots (each equal to
2) and f(1, x) = 0 also has equal roots (each equal to 0). Find the equation of the circle.​

Answer 1

Answer:

f(x,y)=0

f(x,1)=0

***f(x,y)=f(x).f(y)

@ f(x)=0

f(1,x)=0

similarly f(1-x)=f(x-1)

therefore (x-1)(1-x)=0

Answer 2

Answer:

=(x−2)2+y2–1

Step-by-step explanation:

Start with  f(x,y)=(x−a)2+(y−b)2−r2=0  as the equation of the circle.

Then substitute  y=1  to get  (x−a)2+(1−b)2−r2=0 , and compare this with the quadratic  (x−2)2=0  which has equal roots equal to 2. You should be able to find a few relationships between  a ,  b  and  r .

Then substitute  x=1  to get  (1−a)2+(y−b)2−r2=0  and compare this with the quadratic  y2=0  which has equal roots equal to 0. You will find a few other relationships between  a ,  b  and  r ... enough to find these unknowns, and, consequently, the equation of the circle

f(x,y)=(x−a)2+(y−b)2−r2=0

Given  f(x,1)=0

⟹(x−a)2+(−b)2−r2=0

⟹x2–2ax+a2+b2−r2=0

But given  x2–2ax+a2+b2−r2=x2–4x+4=0

⟹a=2  and  a2+b2−r2=4

⟹b2=r2

Now Given  f(1,y)=(1−a)2+(y−b)2−r2=y2

⟹1–2a+a2+y2–2yb+b2−r2=y2

⟹1–2a+a2–2yb+b2−r2=0

But  a=2  and  b2−r2=0

⟹1–4+4–2yb=0

⟹2yb=1

Now consider  f(x,y)=(x−a)2+(y−b)2−r2

=x2–2ax+a2+y2–2yb+b2−r2

=x2–4x+4+y2–1+0

=(x−2)2+y2–1