Find the equation of pair of tangents from
(i) (0,0) to the circle `x^(2)+y^(2)+10x+10y+40=0`
(ii) (4,10) to the circle `x^(2)+y^(2)=25`
(iii) (3,2) to the circle `x^(2)+y^(2)-6x+4y-2=0`
(iv) (10,4) to the circle `x^(2)+y^(2)=25`
(v) (1,3) to the circle `x^(2)+y^(2)-2x+4y-11=0`

Answer

To find the equations of the pair of tangents from a point to a circle, we use the formula \( t^2 = SS_1 \), where \( S \) is the equation of the circle and \( S_1 \) is obtained by substituting the coordinates of the external point into the equation of the circle.

### (i) From (0,0) to the circle \( x^2 + y^2 + 10x + 10y + 40 = 0 \)

1. **Rearranging the circle equation**:
The equation of the circle can be rewritten as:
\[ x^2 + y^2 + 10x + 10y + 40 = 0 \]

2. **Identifying \( S \)**:
Here, \( S = x^2 + y^2 + 10x + 10y + 40 \).

3. **Calculating \( S_1 \)**:
For point \( (0,0) \):
\[ S_1 = 0^2 + 0^2 + 10(0) + 10(0) + 40 = 40 \]

4. **Applying the formula**:
\[ t^2 = SS_1 \implies t^2 = (x^2 + y^2 + 10x + 10y + 40)(40) \]

5. **Expanding and simplifying**:
\[ t^2 = 40(x^2 + y^2 + 10x + 10y + 40) \]

6. **Final equation**:
The equation of the pair of tangents is:
\[ 40x^2 + 40y^2 + 400x + 400y + 1600 = 0 \]