I'll assume the following:
$$\text{Height of cone } = h \\
\text{Radius of the cone at its base } = R \\
\text{Distance from the cone axis to the slicing plane } = d,\quad 0\leq d\leq R.$$
Each horizontal slice of the solid is a circular segment. These segments vary in size due to the change of radius as we move vertically up or down the cone. Let's suppose at any such height $z$ the circle has radius $r$ and the segment subtends an angle $\theta$ at the circle centre (see the picture at the link below). Then, simple trigonometry/geometry shows that
$$\cos\frac{\theta}{2} = \frac{d}{r}, \\
\text{and the thickness of these slices will be: } dz = \dfrac{h}{R}dr.$$
(Actually, $r$ decreases as $z$ increases but this is handled by flipping the limits of integration below on $r$: $d$ to $R$ instead of $R$ to $d$.)
The area for the segment is:
\begin{align}
A_s &= \dfrac{r^2}{2}\left(\theta - \sin\theta\right) \\
&= \dfrac{r^2}{2}\left(2\cos^{-1}\dfrac{d}{r} - \sin\left(2\cos^{-1}\dfrac{d}{r}\right)\right) \\
&= r^2\cos^{-1}\dfrac{d}{r} - d\sqrt{r^2-d^2} \\
& \qquad\text{using $\;\sin(2\alpha)=2\sin\alpha\cos\alpha\;$ and $\;\cos(\sin^{-1}(a/b))=\sqrt{b^2-a^2}/b$}.
\end{align}
Therefore the volume of a thin horizontal slice of thickness $dz\;$ is
\begin{align}
dV &= A_sdz \\
&= A_s\dfrac{h}{R}dr \\
&= \dfrac{h}{R}\left(r^2\cos^{-1}\dfrac{d}{r} - d\sqrt{r^2-d^2}\right)dr.
\end{align}
Therefore the required volume is:
\begin{align}
V &= \int_{r=d}^{R} \dfrac{h}{R}\left(r^2\cos^{-1}\dfrac{d}{r} - d\sqrt{r^2-d^2}\right)\;dr \\
&= \dfrac{h}{3R}\left[-2dr\sqrt{r^2-d^2} + d^3\ln\left(\sqrt{r^2-d^2}+r\right) + r^3\cos^{-1}\dfrac{d}{r} \right]_{r=d}^{R} \\
& \qquad\text{using Wolfram Alpha} \\
&= \dfrac{h}{3}\left[-2d\sqrt{R^2-d^2} + \dfrac{d^3}{R}\ln\left(\sqrt{R^2-d^2}+R\right) + R^2\cos^{-1}\dfrac{d}{R} - \dfrac{d^3}{R}\ln{d} \right]. \\
\end{align}
As a sanity check, setting $d=0$ gives $V=\pi R^2h/6,$ and setting $d=R$ gives $V=0,$ as expected.
We may let $t=\frac{y}{x}$ and solve the equation about $t$:
$$
x^2+y^2 = 2ax + 2by \implies 1+t^2 =\frac2{x}(a+bt),
$$
$$
y=mx+c\implies t=m+\frac{c}x\implies \frac2 x = 2\frac{t-m} c.
$$ Thus
$$
1+t^2 = 2\left(\frac{t-m} c\right)(a+bt),
$$ which is equivalent to
$$
(2b-c)t^2 +2(a-mb)t-(2am+c)=0.\tag {*}
$$ The solution is given by
$$
t_i=\frac{(a-mb)\pm\sqrt{(a-mb)^2+(2b-c)(2am+c)}}{2b-c},\ \ i=1,2,
$$ provided that $D/4 =(a-mb)^2+(2b-c)(2am+c)=-c^2+(2b-2am)c+(a+mb)^2\ge 0$.
These two lines $y=t_i x$ form a right angle at the origin, that is, are orthogonal to each other if and only if $t_1t_2 = -1$. By Vieta's formula, this is equivalent to $\frac{2am+c}{2b-c}=1$, i.e. $b=am+c$. We note that this is also equivalent to that the line $y=mx+c$ passes through the center of the circle $(a,b)$.
The line $y=mx+c$ touches the circle if and only if the equation $(*)$ has multiple roots. i.e. $D/4 =0$. Arranging the terms, we find that it is equivalent to
$$
-(c+am-b)^2 +(a^2+b^2)(m^2+1)= 0\Longleftrightarrow c= b-am\pm \sqrt{(a^2+b^2)(m^2+1)}.
$$ We also note that this is equivalent to
$$
\frac{|b-am-c|}{\sqrt{m^2+1}}=\sqrt{a^2+b^2},
$$ where the LHS is the distance between the line $y-mx-c=0$ and the center of the circle $(a,b)$, and the RHS is the radius of the circle.
Best Answer
You can see here for interactive graph