Question

Find the equation of the circle determined by the given conditions:
1. passing through (1, 2), (2, 3) and (-2, 1)
2. passing through (4, 6), (-2, -2) and (-4, 2)
3. circumscribing a triangle formed by the lines x - y + 2 = 0, 2x + 3y - 1 = 0 and 4x + y + 17 = 0
4. center of the x-axis and passing through (0, 0) and (2, 4)
5. containing the point (2, 2) and tangent to the lines y = 1 and y = 6.
6. tangent to the x-axis and passing through (5, 1) and (-2, 8)
7. passing through (-3, -1), (3, -5) and having its center on the line 2x - y - 2 = 0
8. tangent to the line 4x - 3y = 6 at (3, 2) and passing through (2, -1)
9. tangent to the line 3x - 4y - 5 = 0 at (3, 1) and passing through (-3, -1)

Answer 1

Answer the question #57675 – Mathematics – Analytic Geometry

Find the equation of the circle determined by the given conditions:

1. passing through $(1,2)$, $(2,3)$ and $(-2,1)$

2. passing through $(4,6)$, $(-2,-2)$ and $(-4,2)$

3. circumscribing a triangle formed by the lines $x - y + 2 = 0$, $2x + 3y - 1 = 0$ and $4x + y + 17 = 0$

4. center of the $x$-axis and passing through $(0,0)$ and $(2,4)$

5. containing the point $(2,2)$ and tangent to the lines $y = 1$ and $y = 6$.

6. tangent to the $x$-axis and passing through $(5,1)$ and $(-2,8)$

7. passing through $(-3,-1)$, $(3,-5)$ and having its center on the line $2x - y - 2 = 0$

8. tangent to the line $4x - 3y = 6$ at $(3,2)$ and passing through $(2,-1)$

9. tangent to the line $3x - 4y - 5 = 0$ at $(3,1)$ and passing through $(-3,-1)$

Solutions.

1. So as an equation of any circle is $(x - a)^2 + (y - b)^2 = r^2$ then we have the next system:

$$\left\{ \begin{array}{l} (1 - a)^2 + (2 - b)^2 = r^2, \\ (2 - a)^2 + (3 - b)^2 = r^2, \\ (-2 - a)^2 + (1 - b)^2 = r^2, \end{array} \right. \to \left\{ \begin{array}{l} 1 - 2a + a^2 + 4 - 4b + b^2 = r^2, \\ 4 - 4a + a^2 + 9 - 6b + b^2 = r^2, \\ 4 + 4a + a^2 + 1 - 2b + b^2 = r^2, \end{array} \right. \to \left\{ \begin{array}{l} 2a + 2b = 8, \\ 6a + 2b = 0, \\ 5 - 2a + a^2 - 4b + b^2 = r^2, \end{array} \right. \to \left\{ \begin{array}{l} a = -2, \\ b = 6, \\ r^2 = 25. \end{array} \right.$$

Thus, the equation of the circle determined by the given conditions is $(x + 2)^2 + (y - 6)^2 = 25$.

2. So as an equation of any circle is $(x - a)^2 + (y - b)^2 = r^2$ then we have the next system:

$$\left\{ \begin{array}{l} (4 - a)^2 + (6 - b)^2 = r^2, \\ (-2 - a)^2 + (-2 - b)^2 = r^2, \\ (-4 - a)^2 + (2 - b)^2 = r^2, \end{array} \right. \to \left\{ \begin{array}{l} 16 - 8a + a^2 + 36 - 12b + b^2 = r^2, \\ 4 + 4a + a^2 + 4 + 4b + b^2 = r^2, \\ 16 + 8a + a^2 + 4 - 4b + b^2 = r^2, \end{array} \right. \to \left\{ \begin{array}{l} 3a + 4b = 11, \\ 4b - 2a = 6, \\ 8 + 4a + a^2 + 4b + b^2 = r^2, \end{array} \right. \to \left\{ \begin{array}{l} a = 1, \\ b = 2, \\ r^2 = 25. \end{array} \right.$$

Thus, the equation of the circle determined by the given conditions is $(x - 1)^2 + (y - 2)^2 = 25$.

3. The first we solve the systems to find three points of the circle:

$$\left\{ \begin{array}{l} x - y + 2 = 0, \\ 2x + 3y - 1 = 0, \end{array} \right. \left\{ \begin{array}{l} x = y - 2, \\ 5y - 5 = 0, \end{array} \right. \left\{ \begin{array}{l} x = -1, \\ y = 1, \end{array} \right.$$

We get the point $(-1,1)$:

$$\left\{ \begin{array}{l} x - y + 2 = 0, \\ 4x + y + 17 = 0, \\ 4x + y + 17 = 0, \\ 2x + 3y - 1 = 0, \\ 5y = 19, \\ x = -5.2, \\ y = 1.8, \end{array} \right.$$

we get the point $(-3.8, -1.8)$:

So as an equation of any circle is $(x - a)^2 + (y - b)^2 = r^2$ then we have the next system:

$$\left\{ \begin{array}{l} (-1 - a)^2 + (1 - b)^2 = r^2, \\ (-3.8 - a)^2 + (-1.8 - b)^2 = r^2, \\ (-5.2 - a)^2 + (3.8 - b)^2 = r^2, \end{array} \right. \to \left\{ \begin{array}{l} 1 + 2a + a^2 + 1 - 2b + b^2 = r^2, \\ 14.44 + 7.6a + a^2 + 3.24 + 3.6b + b^2 = r^2, \\ 27.04 + 10.4a + a^2 + 14.44 - 7.6b + b^2 = r^2, \end{array} \right. \to \left\{ \begin{array}{l} a = -3.94, \\ b = 1.14, \\ r^2 = 8.6632. \end{array} \right.$$

Thus, the equation of the circle determined by the given conditions is

$$(x + 3.94)^2 + (y - 1.14)^2 = 8.6632.$$

4. The line, which is passing through $0, 0$ and $B(2, 4)$ is $\frac{x}{2} = \frac{y}{4}$ or $2x - y = 0$ . The midpoint of the segment $AB$ is $M(1, 2)$ and midperpendicular to the line $AB$ is $x - 1 + 2(y - 2) = 0$ or $x + 2y - 5 = 0$ and if the center $a, b$ of the circle $x - a)^2 + y - b)^2 = r^2$ is on the $x$ -axis then the center is the point $(5, 0)$ and its equation is $x - 5)^2 + y^2 = 25$ .

5. If the circle tangent to the lines $y = 1$ and $y = 6$ , then its center is on the line $y = \frac{1 + 6}{2} = 3.5$ and its radius is equal to 2.5. So as the circle is containing the point 2, 2) then we have: $2 - a)^2 + 2 - 3.5)^2 = 2.5^2$ ; $2 - a)^2 = 6.25 - 2.25 = 4$ ; $2 - a = \pm 2$ and $a = 0$ or $a = 4$ . Thus, the equation of the circle determined by the given conditions is $x - 4)^2 + y - 3.5)^2 = 625$ or $x^2 + (y - 3.5)^2 = 625$ .

6. If the circle is passing through the points 5, 1) and $B(-2,8)$ then its center is on the line, which is midperpendicular to the line $AB: \frac{x + 2}{7} = \frac{y - 8}{-7}$ or $x + y - 6 = 0$ . The midpoint of $AB$ is $M(1.5,4.5)$ , then he center is on the line $(x - 1.5) - y + 4.5 = 0$ or $x - y + 3 = 0$ . In other side, if circle tangent to the $x$ -axis the center $a,b$ of this circle must satisfy the condition $b^2 = a - 5)^2 + b - 1)^2$ . We have the next system:

$$\left\{ \begin{array}{c} a - b + 3 = 0, \\ b ^ {2} = a - 5) ^ {2} + b - 1) ^ {2}, \end{array} \right. \to \left\{ \begin{array}{c} b = a + 3, \\ b ^ {2} = a ^ {2} - 1 0 a + 2 5 + b ^ {2} - 2 b + 1, \end{array} \right.$$ $$\left\{ \begin{array}{c} b = a + 3, \\ a ^ {2} - 1 0 a + 2 6 - 2 b = 0, \end{array} \right. \to \left\{ \begin{array}{c} b = a + 3, \\ a ^ {2} - 1 2 a + 2 0 = 0, \end{array} \right. \to \left\{ \begin{array}{c} b = 5, \\ a = 2. \end{array} \right.$$

and the equation of the circle determined by the given conditions is $(x - 2)^{2} + y - 5)^{2} = 25$ .

7. If the circle is passing through the points $-3, -1$ and $B(3, -5)$ then its center is on the line, which is midperpendicular to the line $AB: \frac{x + 3}{6} = \frac{y + 1}{-4}$ or $2x + 3y + 9 = 0$ . The midpoint of $AB$ is $M(0, -3)$ , then he center is on the line $3x - 2(y + 3) = 0$ or $3x - 2y - 6 = 0$ . In other side, the center $a, b$ of this circle is the line $2x - y - 2 = 0$ . We have the next system:

$$\left\{ \begin{array}{l} 3 a - 2 b - 6 = 0, \\ 2 a - b - 2 = 0, \end{array} \right. \to \left\{ \begin{array}{l} a = - 2, \\ b = - 6. \end{array} \right.$$

The square of radius of this circle is equal to $r^2 = -3 + 2)^2 + -1 + 6)^2 = 26$ and the equation of the circle determined by the given conditions is $(x + 2)^2 + y + 6)^2 = 26$ .

8. If the circle is tangent to the line $4x - 3y = 6$ at 3,2) then its center is on the line, which is perpendicular to the given line and is containing the point 3,2). So it is line $3(x - 3) + 4(y - 2) = 0$ or $3x + 4y - 17 = 0$ . In other side, if circle is passing through 2,-1) then the center $a,b$ of this circle must satisfy the condition $a - 3)^2 + b - 2)^2 = a - 2)^2 + b + 1)^2$ . We have the next system:

$$\left\{ \begin{array}{c} 3 a + 4 b - 1 7 = 0, \\ a - 3) ^ {2} + b - 2) ^ {2} = a - 2) ^ {2} + b + 1) ^ {2}, \end{array} \right. \to \left\{ \begin{array}{c} 3 a = 1 7 - 4 b, \\ - 6 a + 9 - 4 b + 4 = - 4 a + 4 + 2 b + 1, \end{array} \right.$$ $$\left\{ \begin{array}{l} 3 a = 1 7 - 4 b, \\ a = 4 - 3 b, \end{array} \right. \to \left\{ \begin{array}{l} b = - 1, \\ a = 7. \end{array} \right.$$

and the equation of the circle determined by the given conditions is $(x - 7)^2 + y + 1)^2 = 25$ .

9. If the circle is tangent to the line $3x - 4y - 5 = 0$ at 3,1) then its center is on the line, which is perpendicular to the given line and is containing the point 3,1). So it is line $4(x - 3) + 3y - 1) = 0$ or $4x + 3y - 15 = 0$ . In other side, if circle is passing through $-3, -1$ then the center $a, b$ of this circle must satisfy the condition $a - 3)^2 + b - 1)^2 = a + 3)^2 + b + 1)^2$ . We have the next system:

$$\left\{ \begin{array}{c} 4 a + 3 b - 1 5 = 0, \\ a - 3) ^ {2} + b - 1) ^ {2} = a + 3) ^ {2} + b + 1) ^ {2}, \end{array} \right. \to \left\{ \begin{array}{c} 4 a = 1 5 - 3 b, \\ - 6 a + 9 - 2 b + 1 = 6 a + 9 + 2 b + 1, \end{array} \right.$$ $$\left\{ \begin{array}{l} 4 a = 1 5 - 3 b, \\ 1 2 a = - 4 b, \end{array} \right. \to \left\{ \begin{array}{l} b = 9, \\ a = - 3. \end{array} \right.$$

and the equation of the circle determined by the given conditions is $(x + 3)^2 + y - 9)^2 = 100$ .

Answer: all sections are solved.

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