Home
Class 12
MATHS
For the circles 3x^2+3y^2+x+2y-1=0 , 2x^...

For the circles `3x^2+3y^2+x+2y-1=0 , 2x^2+2y^2+2x-y-1=0` radical axis is

A

4x-7y=1

B

3x+y-2=0

C

x-3y=0

D

4x-7y=5

Text Solution

AI Generated Solution

The correct Answer is:
To find the radical axis of the given circles, we will follow these steps: ### Step 1: Write the equations of the circles The equations of the circles are given as: 1. \( S_1: 3x^2 + 3y^2 + x + 2y - 1 = 0 \) 2. \( S_2: 2x^2 + 2y^2 + 2x - y - 1 = 0 \) ### Step 2: Simplify the equations We can simplify both equations by dividing them by their respective coefficients. For \( S_1 \): \[ S_1: \frac{3x^2 + 3y^2 + x + 2y - 1}{3} = 0 \implies x^2 + y^2 + \frac{x}{3} + \frac{2y}{3} - \frac{1}{3} = 0 \] For \( S_2 \): \[ S_2: \frac{2x^2 + 2y^2 + 2x - y - 1}{2} = 0 \implies x^2 + y^2 + x - \frac{y}{2} - \frac{1}{2} = 0 \] ### Step 3: Set up the equation for the radical axis The radical axis can be found using the condition \( S_1 - S_2 = 0 \). Subtract \( S_2 \) from \( S_1 \): \[ S_1 - S_2 = \left( x^2 + y^2 + \frac{x}{3} + \frac{2y}{3} - \frac{1}{3} \right) - \left( x^2 + y^2 + x - \frac{y}{2} - \frac{1}{2} \right) = 0 \] ### Step 4: Simplify the equation Now, simplifying the equation: \[ \frac{x}{3} - x + \frac{2y}{3} + \frac{y}{2} - \frac{1}{3} + \frac{1}{2} = 0 \] Combine like terms: - The \( x \) terms: \( \frac{x}{3} - x = \frac{x - 3x}{3} = -\frac{2x}{3} \) - The \( y \) terms: \( \frac{2y}{3} + \frac{y}{2} \) - The constants: \( -\frac{1}{3} + \frac{1}{2} = \frac{-2 + 3}{6} = \frac{1}{6} \) So we have: \[ -\frac{2x}{3} + \left(\frac{2y}{3} + \frac{3y}{6}\right) + \frac{1}{6} = 0 \] Finding a common denominator for \( y \): \[ -\frac{2x}{3} + \left(\frac{4y}{6} + \frac{3y}{6}\right) + \frac{1}{6} = 0 \] \[ -\frac{2x}{3} + \frac{7y}{6} + \frac{1}{6} = 0 \] ### Step 5: Clear the fractions Multiply through by 6 to eliminate the denominators: \[ -4x + 7y + 1 = 0 \] ### Step 6: Rearranging the equation Rearranging gives us the equation of the radical axis: \[ 4x - 7y = -1 \quad \text{or} \quad 4x - 7y = 1 \] ### Final Answer Thus, the equation of the radical axis is: \[ 4x - 7y = 1 \]
Promotional Banner

Similar Questions

Explore conceptually related problems

The intercept made by the radical axis of the circles x^2+y^2+6x-16=0 and x^2+y^2-2x-6y-6=0 on x-axis is

Area of the triangle formed by the radical axis of the circles x^2+y^2=4, x^2+y^2+2x+4y-6=0 with co-ordinate axes is

The radical centre of the circles x^2+y^2=9, x^2+y^2-2x-2y-5=0 , x^2+y^2+4x+6y-19=0 is

If the radical axis of the circles x^2+y^2-1=0 and x^2+y^2-2x-2y+1=0 form the triangle of area A with co-ordinate axes, then the value of 1/A is 1 (2) 2 (3) 3 (4) 4

The length of the common chord of the circles x^(2)+y^(2)-2x-1=0 and x^(2)+y^(2)+4y-1=0 , is

Equation of the circle cutting orthogonal these circles x^2+y^2-2x-3y-7=0 , x^2 +y^2+5x-5y+9=0 and x^2+y^2+7x-9y+29=0 is:

Coordinates of the centre of the circle which bisects the circumferences of the circles x^2 + y^2 = 1; x^2 + y^2 + 2x - 3 = 0 and x^2 + y^2 + 2y-3 = 0 is

The equation of the circle which cuts orthogonally the three circles x^2+y^2+4x+2y+1=0, 2x^2+2y^2+8x+6y-3=0, x^2+y^2+6x-2y-3=0 is

Find the equation of the circle which cuts the circles x^2+y^2+4x+2y+1=0, 2(x^2+y^2)+8x+6y-3=0 and x^2+y^2+6x-2y-3=0 orthogonally.

Find the radical centre of the following circles. x^2 + y^2 + 4x - 7 = 0 , 2x^2 + 2y^2 + 3x + 5y - 9 = 0 , x^2 + y^2 + y = 0 .