[tex]r=\dfrac{10}{10-7\cos\theta}\implies10r-7r\cos\theta=10[/tex]
Converting to Cartesian coordinates using
[tex]x=r\cos\theta[/tex]
[tex]r=\sqrt{x^2+y^2}[/tex]
we have
[tex]10\sqrt{x^2+y^2}-7x=10[/tex]
[tex]10\sqrt{x^2+y^2}=7x+10[/tex]
[tex]100(x^2+y^2)=(7x+10)^2[/tex]
[tex]100x^2+100y^2=49x^2+140x+100[/tex]
[tex]51x^2-140x+100y^2=100[/tex]
Complete both squares to get
[tex]51\left(x-\dfrac{70}{51}\right)^2-\dfrac{4900}{51}+100y^2=100[/tex]
[tex]51\left(x-\dfrac{70}{51}\right)^2+100y^2=\dfrac{10,000}{51}[/tex]
[tex]\dfrac{2601}{10,000}\left(x-\dfrac{70}{51}\right)^2+\dfrac{51}{100}y^2=1[/tex]
This is the equation of an ellipse, so it's either A or C.
The ellipse is centered at [tex]\left(\dfrac{70}{51},0\right)\approx(1.37,0)[/tex], so it falls to the right of the vertical axis, which means the answer is C.
