The length of the tangent from $(1, 1)$ to the circle $2x^2+2y^2+5x+3y+1=0$ is?
- $\sqrt{\dfrac{13}{2}}$
- $3$
- $2$
- $1$
The length of the tangent from (x1, y1) to a circle x^2 + y^2 + gx + fy + c = 0 is sqrt(x1^2 + y1^2 + gx1 + fy1 + c). First, normalize the equation by dividing by 2: x^2 + y^2 + 2.5x + 1.5y + 0.5 = 0. Plugging in (1, 1) gives sqrt(1 + 1 + 2.5 + 1.5 + 0.5) = sqrt(6.5) = sqrt(13/2).
To use the standard tangent length formula, divide the equation 2x squared plus 2y squared plus 5x plus 3y plus 1 equals 0 by 2 to get x squared plus y squared plus (5 divided by 2)x plus (3 divided by 2)y plus (1 divided by 2) equals 0. Substituting the point (1, 1) into the expression x squared plus y squared plus (5 divided by 2)x plus (3 divided by 2)y plus (1 divided by 2) gives 1 plus 1 plus (5 divided by 2) plus (3 divided by 2) plus (1 divided by 2). This simplifies to 2 plus 4 plus (1 divided by 2), which equals 13 divided by 2. The length of the tangent is the square root of this value, resulting in the square root of (13 divided by 2).