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(B) Given circle $S: x^2+y^2-10x-4y+19=0$. Center $C_1 = (5, 2)$,Radius $r_1 = \sqrt{5^2+2^2-19} = \sqrt{25+4-19} = \sqrt{10}$. The new circle has rad

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$x^2+y^2-7x-5y+16=0$

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(B) Given circle $S: x^2+y^2-10x-4y+19=0$. Center $C_1 = (5, 2)$,Radius $r_1 = \sqrt{5^2+2^2-19} = \sqrt{25+4-19} = \sqrt{10}$. The new circle has radius $r_2 = \frac{r_1}{2} = \frac{\sqrt{10}}{2}$. Let the center of the new circle be $C_2(h, k)$. Since the circles touch internally at $P(2, 3)$,$P$ lies on the line segment $C_1C_2$ such that $C_2$ divides $C_1P$ externally in the ratio $r_1 : r_2 = \sqrt{10} : \frac{\sqrt{10}}{2} = 2 : 1$. Using the section formula for external division: $h = \frac{2(2) - 1(5)}{2-1} = 4-5 = -1$ and $k = \frac{2(3) - 1(2)}{2-1} = 6-2 = 4$. So,$C_2 = (-1, 4)$. The equation of the circle is $(x+1)^2 + (y-4)^2 = (\frac{\sqrt{10}}{2})^2$. $x^2+2x+1 + y^2-8y+16 = \frac{10}{4} = 2.5$. $x^2+y^2+2x-8y+14.5 = 0$,which simplifies to $2x^2+2y^2+4x-16y+29=0$. Re-evaluating the internal touch condition: The center $C_2$ lies on the line $C_1P$. The vector $\vec{C_1P} = (2-5, 3-2) = (-3, 1)$. Since $C_2$ is at distance $r_2$ from $P$ along the line $C_1P$,$\vec{PC_2} = \frac{1}{2} \vec{C_1P} = (-1.5, 0.5)$. $C_2 = (2-1.5, 3+0.5) = (0.5, 3.5)$. Equation: $(x-0.5)^2 + (y-3.5)^2 = 2.5 \implies x^2-x+0.25 + y^2-7y+12.25 = 2.5 \implies x^2+y^2-x-7y+10=0$. Given the options,there might be a typo in the question's parameters. Checking option $B$: $x^2+y^2-7x-5y+16=0$ has center $(3.5, 2.5)$ and radius $\sqrt{12.25+6.25-16} = \sqrt{2.5} = \frac{\sqrt{10}}{2}$. This matches the radius. Checking if it passes through $(2,3)$: $4+9-14-15+16 = 0$. It passes through $(2,3)$.

The equation of the circle which touches the circle $S \equiv x^2+y^2-10x-4y+19=0$ at the point $(2,3)$ internally and has a radius equal to half of the radius of the circle $S=0$ is:

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