Question

Find the equation of circle with center at the origin and tangent to the line 2x - 5y =8.

Answer 1

Find the equation of circle with center at the origin and tangent to the line $2x - 5y = 8$.

**Solution:**

Equation of circle with center at the origin $(0,0)$ and radius $r$: $x^{2} + y^{2} = r^{2}$, $r > 0$.

Points of intersection of the circle and a line:

$$\left\{ \begin{array}{l} 2x - 5y = 8 \\ x^{2} + y^{2} = r^{2} \end{array} \right.$$ $$\left\{ \begin{array}{l} x = \frac{1}{2}(5y + 8) \\ x^{2} + y^{2} = r^{2} \end{array} \right.$$ $$\frac{1}{4}(5y + 8)^{2} + y^{2} = r^{2}$$ $$25y^{2} + 80y + 64 + 4y^{2} - 4r^{2} = 0$$ $$29y^{2} + 80y + (64 - 4r^{2}) = 0$$ $$D = 80^{2} - 4 \cdot 29 \cdot (64 - 4r^{2}) = 6400 - 7424 + 464r^{2} = 464r^{2} - 1024$$

Circle is tangent to line $\Longleftrightarrow$ Circle and line have only one intersection $\Longleftrightarrow D = 0$.

$$D = 464r^{2} - 1024 = 0$$ $$r^{2} = \frac{1024}{464} = \frac{64}{29}$$

Equation of circle: $x^{2} + y^{2} = r^{2} = \frac{64}{29}$

Answer: $x^{2} + y^{2} = \frac{64}{29}$