The number of real values of $t\in\mathbb{R}$ for which exactly one circle passing through the points $(2,-3)$ and $(t,2t-1)$ and touching the line $16x-2y+27=0$ is
Here the point $(t,2t-1)$ always lie in line $y=2x-1\implies 2x-y-1=0$
Now let $(h,k)$ be the center of circle, Then using distance formula
$\displaystyle \sqrt{(h-2)^2+(k+3)^2}=\frac{|16h-2k+26|}{\sqrt{16^2+2^2}}=\frac{|2h-k-1|}{\sqrt{2^2+1^2}}\cdots (1)$
From $\displaystyle 16h-2k+26=52(2h-k-1)$
$88h-50k-78=0\implies 44h-25k-39=0$
Now equating $k$ from above equation in $(1)$ St equation
But this is very lengthy way.
Please have a look, how can I go ahead?
Let $A$ be the point on line $=2−1$, and $B=(2,-3)$. Let $F$ be the point where line $AB$ meets line $r$ of equation $16−2+27=0$. If $A$ and $B$ are on the same side of $r$, then a circle $ABT$ exists, tangent to $r$ at $T$. And if $T'$ is the reflection of $T$ about $F$, then circle $ABT'$ is also tangent to $r$, due to the properties of radical axis $AB$.
Hence, if $A$ and $B$ are on the same side of $r$, we always get two tangent circles, whatever the position of $A$, unless $AB$ is parallel to $r$. And that only happens for $t=3$.
If $A$ lies on $r$ ($t=-29/12$), then $A=F=T=$ intersection point of the lines and we also have a single circle, tangent to $r$ at $A$. Finally, if $A$ and $B$ are on opposite sides of $r$, no tangent circle exists.
In summary, there are only $2$ real values of $t$ for which exactly one circle through $A$, $B$ and touching $r$ exists.