Question

find the equation of this circle

Answer 1

Let the equation of the circle be-

(x-h)² + (y-k)² = r², ...........(i)

where (h,k) are the coordinates of origin and r = length of radius.

Now,

The circle passes through (0, 1).

Thus,

(i) ⇒ (0 - h)² + (1 - k)² = r²

⇒ h² + k² -2k +1 = r² ...........(ii)

Also, it passes through (1, 0)

(i) ⇒ (1 - h)² + (0 - k)² = r²

⇒ h² -2h +1 + k² = r² ...............(iii)

It passes through (1, 1)

(i)⇒ (1-h)² + (1-k)² = r²

⇒ h² -2h +1 + k² -2k + 1 = r²

⇒ h² + k² -2h -2k +2 = r² .............(iv)

Now,

From (ii) and (iii),

h² + k² -2k + 1 = h² + k² -2h +1

⇒ 2k = 2h

⇒ k = h

Now,

(iv) ⇒ h² + h² -2h - 2h +2 = r²

⇒ 2h² -4h +2= r² ...........(v)

Now,

(ii)⇒ h² + h² - 2h + 1 = 2h² -4h + 2

⇒ 2h² -2h +1 = 2h² - 4h + 2

⇒-2h +1 = -4h + 2

⇒ 2h = 1

∴ k = 1/2

(v)⇒ 2h² + 4h = r²

⇒ 2(1/2)² + 4(1/2) = r²

⇒ 1/2 + 2 = r²

⇒ 5/2 = r²

Now,

(i) ⇒ ( x - h)² + (y - k)² = r²

⇒ (x+1/2)² + (y+1/2)² = 5/2

⇒ x² + x + 1/4 + y² + y + 1/4 = 5/2

⇒ 4x²+4x+1+4y²+4y+1/4 = 5/2

⇒ 4x² + 4(x+y) +2 = 20/2

⇒ 4x² +4(x+y) +2 = 10

⇒2x² +2(x+y) +1 = 5

⇒ 2x² + 2(x+y) -4 = 0

⇒ x² + x + y - 2 = 0

Which is the reqd equation of the circle.

Answer 2

Required Solution :-

Here we need to find out the equation of a circle and that circle is passing out to three points (Given in question).

So we would have two methods to find out either by using standard equation of circle or general equation of circle. We would be using "General equation of circle" to do so to prevent complex calculations.

★ General equation of circle :-

• x² + y² + 2gx + 2fy + c = 0

Here,

• g and after are the coordinates of centre

Now, we would be using this equation of circle and forming out three equations. By solving those three equations we would get the values of g, f , and c.

★ Substituting (0 , 1) :

→ (0)² + (1)² + 2g(0) + 2f(1) + c = 0

→ 0 + 1 + 0 + 2f + c = 0

→ 1 + 2f + c = 0 ----- [Equation No.1]

★ Substituting (1 , 0) :

→ (1)² + (0)² + 2g(1) + 2f(0) + c = 0

→ 1 + 0 + 2g + 0 + c = 0

→ 1 + 2g + c = 0 -----[Equation No.2]

★ Substituting (1 , 1) :

→ (1)² + (1)² + 2g(1) + 2f(1) + c = 0

→ 1 + 1 + 2g + 2f + c = 0

→ 2 + 2g + 2f + c = 0 -----[Equation No.3]

★ Solving Equation No.1 & Equation No.3 :

→ 2f = -1 - c

Now,

→ 2 + 2g + (-1 - c) + c = 0

→ 2 + 2g - 1 - c + c = 0

→ 2 + 2g - 1 = 0

→ 2g + 1 = 0

→ g = -½

★ Substituting value of g in Equation No. 2 :

→ 1 + 2(-½) + c = 0

→ 1 - 1 + c = 0

→ c = 0

★ Substituting value of c in Equation No. 1 :

→ 1 + 2f + 0 = 0

→ 1 + 2f = 0

→ 2f = -1

→ f = -½

★ Now, as we got the values of g , f and c so we would substitute it into General equation of circle :

→ x² + y² + 2 (-½) x + 2 (-½) + 0 = 0

→ x² + y² - x - y = 0