If the length of tangent from $(1,1)$ to the circle $2x^2+2y^2-4x-6y+k=0$ is 5 units, find k

Question

The centre of the circle is (1,3/2). The distance from center to (1,1) is 1/2 unit.

The length of tangent is 5 unit. How can it be greater than the distance from the center, since it forms a right angled triangle with the radius.

Answer 1

The length of the tangent from the point $(h,k)$ is $L=\sqrt{F(x,y)}$ if the coefficients of $x^2, y^2$ are 1. Then, $$L(h,k)=\sqrt{h^2+k^2-2h-3k+K/2}\implies \sqrt{K/2-3}=5 \implies K=56$$

But finally the equation $x^2+y^2-2x-3y+28$ does not represent a circle as its radius $r=\sqrt{1+9/4-28}$ becomes imaginary.

We can state that the Eq. $x^2+y^2-2x-3y+K/2=0$ can represent a circle! only if $K<13/2$. Next, for tangents from $(1,1)$ to be possible (for the point $(1,1)$ to be outside the circle)$, L=\sqrt{1+1-2-3+K/2}>0$, this put a further restriction on $K$ as $K<6.$

Finally, The correct answer to OP's question is no real value of $K$ is possible any real value found carelessly (Like my earlier answer 56) will attract inconsistencies.