A circle passes through (- 2, 4) and touches the y - axis at (0, 2)...

As $y$-axis is tangent to the circle therefore the centre of the circle would be $\left(x, 2\right)$ and distance of centre from~\left(0, 2\right) and \left(- 2, 4\right)~should be equal.

$∴\sqrt{h^{2} + {0}^{2}} =\sqrt{\left(h + 2\right)2 + {2}^{2}}$$<br /><br />$⇒h^{2} = h^{2} + 4h + 4 + 4$<br /><br />$⇒h = - 2<br /><br />Hence, centre~is& \left(- 2, 2\right).<br /><br />∴2x - 3y + 10 = 0is& the diameter, as centre lies on it.</p>