Question

Find the standard equation of the circle touching the line x + 2y = 8 at (0, 4) and passing through (3, 7).

Answer 1

Answer on Question #59185 – Math – Analytic Geometry

Question

Find the standard equation of the circle touching the line $x + 2y = 8$ at $(0, 4)$ and passing through $(3, 7)$.

Solution

Since the circle passes through both points A (0, 4) and B (3, 7), the distance between each point and the center of the circle O is equal to

$$\mathrm{AO} = \mathrm{BO} = \mathrm{R};$$ $$(x_0 - x_A)^2 + (y_0 - y_A)^2 = (x_0 - x_B)^2 + (y_0 - y_B)^2 = \mathrm{R}^2;$$ $$(x_0 - 0)^2 + (y_0 - 4)^2 = (x_0 - 3)^2 + (y_0 - 7)^2;$$ $$6x_0 + 6y_0 - 42 = 0;$$ $$x_0 + y_0 - 7 = 0.$$

Therefore, the center of the circle belongs to the line $x + y - 7 = 0$.

On the other hand, since the circle is touching the line $x + 2y = 8$ at A (0, 4), the radius of this circle OA is perpendicular to the line $x + 2y = 8$ and lies on the line b. The equation of line b can be found as follows.

The equation of the line $x + 2y = 8$ can be rewritten as

$$y = \frac{8 - x}{2} = -\frac{x}{2} + 4.$$

Its slope is $k = -0.5$.

The slope of the line b perpendicular to the line $x + 2y = 8$ is

$$k_b = \frac{-1}{k} = \frac{-1}{-0.5} = 2.$$

The equation of line b:

$$y = 2x + b.$$

Since the line $b$ passes through A (0, 4), the intercept $b$ can be found by substitution of the coordinates of point A into the previous equation:

$$y _ {A} = 2 x _ {A} + b.$$ $$b = y _ {A} - 2 x _ {A} = 4 - 2 \times 0 = 4.$$

Therefore, the equation of line $b$ is

$$y = 2 x + 4.$$

Since the center of the circle belongs both to lines

$$y = 2 x + 4$$

and

$$x + y - 7 = 0,$$

it can be found from the following system:

$$\left\{ \begin{array}{l} x + y - 7 = 0 \\ y = 2 x + 4 \end{array} \right.$$

Substituting $y = 2x + 4$ into the first equation of the system yields

$$x + 2x + 4 - 7 = 0;$$ $$3x - 3 = 0;$$ $$x = 1,$$

hence

$$y = 2x + 4 = 2 \times 1 + 4 = 6.$$

Therefore, the center of the circle is the point O(1, 6).

The radius of the circle:

$$R = \sqrt {\left(x _ {0} - x _ {A}\right) ^ {2} + \left(y _ {0} - y _ {A}\right) ^ {2}} = \sqrt {(1 - 0) ^ {2} + (6 - 4) ^ {2}} = \sqrt {5}.$$

The standard equation of the circle is

$$(x - x _ {0}) ^ {2} + (y - y _ {0}) ^ {2} = R ^ {2};$$ $$(x - 1) ^ {2} + (y - 6) ^ {2} = (\sqrt {5}) ^ {2};$$ $$(x - 1) ^ {2} + (y - 6) ^ {2} = 5.$$

Answer:

$$(x - 1) ^ {2} + (y - 6) ^ {2} = 5.$$

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