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Equation of circle passing through (1,sq...

Equation of circle passing through `(1,sqrt(3)), (1,-sqrt(3))` and `(3,-sqrt(3))` is

A

`(x-2)^(2)+y^(2)=4`

B

`(x+2)^(2)+y^(2)=4`

C

`(x-2)^(2)+y^(2)=2`

D

`x^(2)+(y-2)^(2)=4`

Text Solution

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The correct Answer is:
To find the equation of the circle passing through the points (1, √3), (1, -√3), and (3, -√3), we can follow these steps: ### Step 1: General Equation of the Circle The general equation of a circle can be expressed as: \[ x^2 + y^2 + 2gx + 2fy + c = 0 \] where \( g \), \( f \), and \( c \) are constants that we need to determine. ### Step 2: Substitute the First Point (1, √3) Substituting the point (1, √3) into the equation: \[ 1^2 + (\sqrt{3})^2 + 2g(1) + 2f(\sqrt{3}) + c = 0 \] This simplifies to: \[ 1 + 3 + 2g + 2\sqrt{3}f + c = 0 \] \[ 4 + 2g + 2\sqrt{3}f + c = 0 \quad \text{(Equation 1)} \] ### Step 3: Substitute the Second Point (1, -√3) Now, substituting the point (1, -√3): \[ 1^2 + (-\sqrt{3})^2 + 2g(1) + 2f(-\sqrt{3}) + c = 0 \] This simplifies to: \[ 1 + 3 + 2g - 2\sqrt{3}f + c = 0 \] \[ 4 + 2g - 2\sqrt{3}f + c = 0 \quad \text{(Equation 2)} \] ### Step 4: Subtract Equation 2 from Equation 1 Subtracting Equation 2 from Equation 1: \[ (4 + 2g + 2\sqrt{3}f + c) - (4 + 2g - 2\sqrt{3}f + c) = 0 \] This simplifies to: \[ 4\sqrt{3}f = 0 \] Thus, we find: \[ f = 0 \] ### Step 5: Substitute \( f = 0 \) into Equation 1 Substituting \( f = 0 \) into Equation 1: \[ 4 + 2g + c = 0 \quad \text{(Equation 3)} \] ### Step 6: Substitute the Third Point (3, -√3) Now, substituting the point (3, -√3): \[ 3^2 + (-\sqrt{3})^2 + 2g(3) + 2f(-\sqrt{3}) + c = 0 \] This simplifies to: \[ 9 + 3 + 6g + c = 0 \] \[ 12 + 6g + c = 0 \quad \text{(Equation 4)} \] ### Step 7: Substitute \( c \) from Equation 3 into Equation 4 From Equation 3, we have \( c = -4 - 2g \). Substitute this into Equation 4: \[ 12 + 6g + (-4 - 2g) = 0 \] This simplifies to: \[ 12 - 4 + 4g = 0 \] \[ 8 + 4g = 0 \] Thus, we find: \[ g = -2 \] ### Step 8: Find \( c \) Substituting \( g = -2 \) back into Equation 3: \[ 4 + 2(-2) + c = 0 \] \[ 4 - 4 + c = 0 \implies c = 0 \] ### Step 9: Write the Final Equation of the Circle Now we have: - \( g = -2 \) - \( f = 0 \) - \( c = 0 \) Substituting these values into the general equation: \[ x^2 + y^2 + 2(-2)x + 2(0)y + 0 = 0 \] This simplifies to: \[ x^2 + y^2 - 4x = 0 \] ### Step 10: Rearranging the Equation Rearranging gives: \[ x^2 - 4x + y^2 = 0 \] Completing the square for \( x \): \[ (x - 2)^2 - 4 + y^2 = 0 \implies (x - 2)^2 + y^2 = 4 \] ### Final Answer The equation of the circle is: \[ (x - 2)^2 + y^2 = 4 \]
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