Question

Center of the circle (0,0) is tangent to the line x-3y =6

Answer 1

Answer on Question #77927 – Math – Analytic Geometry

Question

Center of the circle $(0,0)$ is tangent to the line $x-3y = 6$

Solution

If the line is tangent to the circle, they have only one common point, and system $\left\{ \begin{array}{l} x^2 + y^2 = R^2 \\ x - 3y = 6 \end{array} \right.$, where $x^2 + y^2 = R^2$ is an equation of a circle with center at $(0,0)$ and radius $R$, has only one real root $(x_0, y_0)$.

$$\left\{ \begin{array}{c} x = 3y + 6 \\ (3y + 6)^2 + y^2 = R^2 \end{array} ; \left\{ \begin{array}{c} x = 3y + 6 \\ 9y^2 + 36y + 36 + y^2 = R^2 \end{array} ; \left\{ \begin{array}{c} x = 3y + 6 \\ 10y^2 + 36y + 36 - R^2 = 0 \end{array} \right. \right. \right.$$

And if the system has only one real root, equation $10y^2 + 36y + (36 - R^2) = 0$ must have only one real root too. Its discriminant is $D = b^2 - 4ac = 36^2 - 4 \times 10 \times (36 - R^2) = 1296 - 1440 + 40R^2$. But if square equation has only one real root – its discriminant is 0

$$1296 - 1440 + 40R^2 = 0;$$ $$40R^2 = 144$$ $$R^2 = 3.6$$

So, the equation of a circle is $x^2 + y^2 = 3.6$. From the equation $10y^2 + 36y + (36 - R^2) = 0$ we can find $y_0 = -\frac{b}{2a} = -\frac{36}{2 \times 10} = -\frac{36}{20} = -1.8$. And from the equation $x = 3y + 6$ we can find $x_0 = 3y_0 + 6 = 3 \times (-1.8) + 6 = -5.4 + 6 = 0.6$.

**Answer**: Equation of a circle is $x^2 + y^2 = 3.6$, and point of contact is $(0.6, -1.8)$.

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