OE


Equation of a circle, having origin as its centre, is in the form 𝑅^2 = 𝑥^2 + 𝑦^2

Given equation of circle is 𝑥^2 + 𝑦^2 = 25,
So, radius of circle is 𝑅 = 5 units. So, circle would be as shown in the figure below.

Now line −3𝑦 = 12 + 2𝑥. Let’s express it in the slope-intercept form, we have \(y=-\frac{2}{3} x-4\)

Here slope is negative and 𝑦 intercept is -4. And 𝑥 intercept is -12.


This is a falling line, crossing 𝑦-axis below 𝑥 −axis at point (0, -4). And crossing 𝑥 −axis at (-12,0)

So, the line is crossing the circle in quadrant III and IV.
Ans. (C and D)