The length of the tangent drawn from the point $(2,3)$ to the circle $2({ x }^{ 2 }+{ y }^{ 2 })-7x+9y-11=0$.
- $18$
- $14$
- $\sqrt{14}$
- $\sqrt{28}$
First, normalize the circle equation: x^2 + y^2 - 3.5x + 4.5y - 5.5 = 0. The length of the tangent from (x1, y1) is sqrt(x1^2 + y1^2 - 3.5x1 + 4.5y1 - 5.5). Substituting (2, 3): sqrt(4 + 9 - 7 + 13.5 - 5.5) = sqrt(14).
The length of the tangent from a point to a circle is given by the formula the square root of the quantity (x1 squared plus y1 squared plus D times x1 plus E times y1 plus F) for a standard circle equation. Dividing the given equation 2(x squared plus y squared) minus 7x plus 9y minus 11 equals 0 by 2 yields x squared plus y squared minus (7 divided by 2)x plus (9 divided by 2)y minus (11 divided by 2) equals 0. Substituting the point (2, 3) into the expression x squared plus y squared minus (7 divided by 2)x plus (9 divided by 2)y minus (11 divided by 2), we get 4 plus 9 minus 7 plus (27 divided by 2) minus (11 divided by 2), which equals 14. Taking the square root of 14 gives the final length.