Home
Class 12
MATHS
If the length of the tangent from (f,g) ...

If the length of the tangent from (f,g) to the circle `x^(2)+y^(2)=6` be twice the length of the tangent from the same point to the circle `x^(2)+y^(2)+3x+3y=0, ` then

A

`x^(2)+y^(2)+4x+4y+2=0`

B

`x^(2)+y^(2)+4x-4y+2=0`

C

`x^(2)+y^(2)-4x-4y+2=0`

D

`x^(2)+y^(2)-4x-4y+2=0`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the locus of the point (f, g) given the conditions about the lengths of tangents to two circles. ### Step-by-Step Solution: 1. **Identify the circles**: - The first circle is given by the equation \( x^2 + y^2 = 6 \). - The second circle is given by the equation \( x^2 + y^2 + 3x + 3y = 0 \). We can rewrite this as \( x^2 + y^2 + 3x + 3y + 0 = 0 \). 2. **Calculate the lengths of the tangents**: - For the first circle \( x^2 + y^2 - 6 = 0 \): - The length of the tangent from the point \( (f, g) \) is given by: \[ l_1 = \sqrt{f^2 + g^2 - 6} \] - For the second circle \( x^2 + y^2 + 3x + 3y = 0 \): - The length of the tangent from the point \( (f, g) \) is given by: \[ l_2 = \sqrt{f^2 + g^2 + 3f + 3g} \] 3. **Set up the relationship between the lengths of the tangents**: - According to the problem, the length of the tangent to the first circle is twice the length of the tangent to the second circle: \[ l_1 = 2l_2 \] 4. **Substituting the expressions for \( l_1 \) and \( l_2 \)**: - Substitute the expressions we found: \[ \sqrt{f^2 + g^2 - 6} = 2\sqrt{f^2 + g^2 + 3f + 3g} \] 5. **Square both sides to eliminate the square roots**: - Squaring both sides gives: \[ f^2 + g^2 - 6 = 4(f^2 + g^2 + 3f + 3g) \] 6. **Expand and rearrange the equation**: - Expanding the right side: \[ f^2 + g^2 - 6 = 4f^2 + 4g^2 + 12f + 12g \] - Rearranging gives: \[ f^2 + g^2 - 4f^2 - 4g^2 - 12f - 12g + 6 = 0 \] - This simplifies to: \[ -3f^2 - 3g^2 - 12f - 12g + 6 = 0 \] 7. **Factor out the common term**: - Dividing the entire equation by -3 gives: \[ f^2 + g^2 + 4f + 4g - 2 = 0 \] 8. **Rewrite the equation**: - This can be rewritten as: \[ f^2 + g^2 + 4f + 4g + 2 = 0 \] 9. **Substituting back to x and y**: - Let \( f = x \) and \( g = y \): \[ x^2 + y^2 + 4x + 4y + 2 = 0 \] ### Final Answer: The locus of the point \( (f, g) \) is given by the equation: \[ x^2 + y^2 + 4x + 4y + 2 = 0 \]
Promotional Banner

Similar Questions

Explore conceptually related problems

If the length of the tangent from (h,k) to the circle x^(2)+y^2=16 is twice the length of the tangent from the same point to the circle x^(2)+y^(2)+2x+2y=0 , then

If the length of the tangent drawn from (f, g) to the circle x^2+y^2= 6 be twice the length of the tangent drawn from the same point to the circle x^2 + y^2 + 3 (x + y) = 0 then show that g^2 +f^2 + 4g + 4f+ 2 = 0 .

If the length of the tangent from a point (f,g) to the circle x^2+y^2=4 be four times the length of the tangent from it to the circle x^(2)+y^(2)=4x , show that 15f^(2)+15g^(2)-64f+4=0

The length of the tangent from (1,1) to the circle 2x^(2)+2y^(2)+5x+3y+1=0 is

If the length of the tangent from (5,4) to the circle x^(2) + y^(2) + 2ky = 0 is 1 then find k.

The length of the tangent from (0, 0) to the circle 2(x^(2)+y^(2))+x-y+5=0 , is

Show that if the length of the tangent from a point P to the circle x^2 + y^2 = a^2 be four times the length of the tangent from it to the circle (x-a)^2 + y^2 = a^2 , then P lies on the circle 15x^2 + 15y^2 - 32ax + a^2=0 .

Find the equations of the tangents from the point A(3,2) to the circle x^(2)+y^(2)+4x+6y+8=0 .

Find the length of the tangents drawn from the point (3,-4) to the circle 2x^(2)+2y^(2)-7x-9y-13=0 .

If the length of tangent drawn from the point (5,3) to the circle x^2+y^2+2x+ky+17=0 is 7, then k= ?