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As you are asking for insights, here is a method able to give you a graphical approximate answer.

Its principle relies on the fact that the set of points $M$ such that

$$MC + \frac{1}{2}MA=c$$

(where $c$ is a constant) is a closed curve that can be called a "generalized ellipse". We have featured some of these curves on the following figure. Their increasing size is in connection with increasing values of constant $c$ (here ranging from $3$ to $6.4$ with $0.2$ steps). The largest value corresponding to a case of tangency with the circle in a point $P$ which is the desired point. Graphicaly one obtains an optimal value

$$c \approx 6.4 \text{indeed very close to } \ \sqrt{41} \ (\approx 6.403...)$$

I think it is possible to use this method in a rigorous way by expressing the proportionality of gradients $G_1$ and $G_2$ in the optimal point of contact :

$$\frac{\vec{PA}}{\|PA\|}+\tfrac12 \frac{\vec{PC}}{\|PC\|}= \lambda \vec{PB}$$

(where $\lambda$ is a Lagrange multiplier) but till now, I haven't succeeded.

As you are asking for insights, here is a method able to give you a graphical approximate answer.

Its principle relies on the fact that the set of points $M$ such that

$$MC + \frac{1}{2}MA=c$$

(where $c$ is a constant) is a closed curve that can be called a "generalized ellipse". We have featured some of these curves on the following figure, here ranging from $c=3$, the smallest, to $c=6.4$ the largest (with $0.2$ steps). This largest value corresponds to a case of approximate tangency with the circle in a point $P$ which is the desired point. This "numerically optimal" value

$$c \ \approx \ 6.4 \ \text{is indeed very close to } \ \sqrt{41} \ (\approx 6.403...)$$

It is probably possible to use this method to derive the exact value in a rigorous way by expressing the proportionality of gradients $G_1$ and $G_2$ in the optimal point of contact :

$$\frac{\vec{PA}}{\|PA\|}+\tfrac12 \frac{\vec{PC}}{\|PC\|}= \lambda \vec{PB}$$

(where $\lambda$ is a Lagrange multiplier) but till now, I haven't succeeded.

Moreover, it is difficult to compete with the nice solution of @Oth S...

As you are asking for insights, here is a method able to give you a graphical approximate answer.

Its priciple is to use the fact that the set of points $M$ such that

$$MC + \frac{1}{2}MA=c$$

(where $c$ is a constant) is a closed curve that can be called a "generalized ellipse". We have featured some of these curves on the following figure. Their increasing size is in correspondance with increasing values of constant $c$ (here ranging from $3 to 4.4$ with $0.2$ steps). The largest value corresponding to a case of tangency with the circle in a point $P$ which is the desired point. The desired optimal value is

$$c \approx 4.4 \ne \sqrt{41}\approx 6.4$$

I think it is possible to use this method in a rigorous way by expressing the proportionality of gradients $G_1$ and $G_2$ in the optimal point of contact :

$$\frac{\vec{PA}}{\|PA\|}+\tfrac12 \frac{\vec{PC}}{\|PC\|}= \lambda \vec{PB}$$

(where $\lambda$ is a Lagrange multiplier) but till now, I haven't succeded.

As you are asking for insights, here is a method able to give you a graphical approximate answer.

Its principle relies on the fact that the set of points $M$ such that

$$MC + \frac{1}{2}MA=c$$

(where $c$ is a constant) is a closed curve that can be called a "generalized ellipse". We have featured some of these curves on the following figure. Their increasing size is in connection with increasing values of constant $c$ (here ranging from $3$ to $6.4$ with $0.2$ steps). The largest value corresponding to a case of tangency with the circle in a point $P$ which is the desired point. Graphicaly one obtains an optimal value

$$c \approx 6.4 \text{indeed very close to } \ \sqrt{41} \ (\approx 6.403...)$$

I think it is possible to use this method in a rigorous way by expressing the proportionality of gradients $G_1$ and $G_2$ in the optimal point of contact :

$$\frac{\vec{PA}}{\|PA\|}+\tfrac12 \frac{\vec{PC}}{\|PC\|}= \lambda \vec{PB}$$

(where $\lambda$ is a Lagrange multiplier) but till now, I haven't succeeded.

As you are asking for insights, here is one that I would qualify as "qualitative", taking into account the fact that the set of points $M$ such that

$$MC + \frac{1}{2}MA=c$$

where $c$ is a constant is a closed curve that can be called a "generalized ellipse". We have featured some of them on the following figure, the largest one being tangent to the circle in the desired optimal point $P$.

I thik it is possible to turn this into a quantitative result by expressing the proportionality of gradients $G_1$ and $G_2$ in the optimal point of contact :

$$\frac{\vec{PA}}{\|PA\|}+\tfrac12 \frac{\vec{PC}}{\|PC\|}= \lambda \vec{PB}$$

(where $\lambda^is a Lagrange mutiplier) but till now, I haven't succeded.

As you are asking for insights, here is a method able to give you a graphical approximate answer.

Its priciple is to use the fact that the set of points $M$ such that

$$MC + \frac{1}{2}MA=c$$

(where $c$ is a constant) is a closed curve that can be called a "generalized ellipse". We have featured some of these curves on the following figure. Their increasing size is in correspondance with increasing values of constant $c$ (here ranging from $3 to 4.4$ with $0.2$ steps). The largest value corresponding to a case of tangency with the circle in a point $P$ which is the desired point. The desired optimal value is

$$c \approx 4.4 \ne \sqrt{41}\approx 6.4$$

I think it is possible to use this method in a rigorous way by expressing the proportionality of gradients $G_1$ and $G_2$ in the optimal point of contact :

$$\frac{\vec{PA}}{\|PA\|}+\tfrac12 \frac{\vec{PC}}{\|PC\|}= \lambda \vec{PB}$$

(where $\lambda$ is a Lagrange multiplier) but till now, I haven't succeded.