Writing Equations from Geometry - Math 3

1. Circle $C_1$ has equation $(x - 2)^2 + (y + 3)^2 = 16$ and circle $C_2$ has equation $(x + 1)^2 + (y - 1)^2 = 9$. What is the equation of the line containing the centers of both circles?
   Answer: $4x + 3y = -1$ The center of circle C₁ is (2, -3) and the center of circle C₂ is (-1, 1). The line through these centers has slope (1 - (-3))/(-1 - 2) = 4/(-3) = -4/3. Using point-slope form with center (2, -3): y - (-3) = (-4/3)(x - 2), so y + 3 = (-4/3)x + 8/3, giving y = (-4/3)x + 8/3 - 3 = (-4/3)x - 1/3. Converting to standard form: 3y = -4x - 1, so 4x + 3y = -1.
2. The center of a circle lies on the line $2x + y = 10$, and the circle is tangent to both the $x$-axis and the line $y = 4$. What is the equation of this circle?
   Answer: $(x - 4)^2 + (y - 2)^2 = 4$ Since the circle is tangent to both the $x$-axis and the line $y = 4$, the center must be equidistant from both lines. The distance from a point $(h, k)$ to the $x$-axis is $|k|$, and the distance to the line $y = 4$ is $|k - 4|$. For tangency to both: $|k| = |k - 4|$. This gives $k = k - 4$ (impossible) or $k = -(k - 4) = -k + 4$, so $2k = 4$ and $k = 2$. The radius is $r = 2$. Since the center lies on $2x + y = 10$: $2h + 2 = 10$, so $h = 4$. Therefore, the center is $(4, 2)$ and the equation is $(x - 4)^2 + (y - 2)^2 = 4$. Choice A has the wrong center. Choice C has $k = 6$, which would make the circle intersect the $x$-axis, not be tangent to it. Choice D has the center on the $x$-axis, making it impossible to be tangent to both lines with the same radius.
3. A rectangle has vertices at $A(1, 2)$, $B(7, 4)$, $C(6, 7)$, and $D(0, 5)$. What is the equation of the circle that passes through all four vertices of this rectangle?
   Answer: $(x - 3.5)^2 + (y - 4.5)^2 = 12.5$ When a rectangle is inscribed in a circle, the circle passes through all four vertices, and its center is at the intersection of the rectangle's diagonals. Since a rectangle's diagonals bisect each other, you can find the circle's center by finding the midpoint of either diagonal. Let's find the center using diagonal $AC$. The midpoint formula gives us: Center = $\left(\frac{1+6}{2}, \frac{2+7}{2}\right) = (3.5, 4.5)$ We can verify this is correct by checking diagonal $BD$: Midpoint = $\left(\frac{7+0}{2}, \frac{4+5}{2}\right) = (3.5, 4.5)$ ✓ Now we need the radius. Calculate the distance from the center $(3.5, 4.5)$ to any vertex, say $A(1, 2)$: $r^2 = (1-3.5)^2 + (2-4.5)^2 = (-2.5)^2 + (-2.5)^2 = 6.25 + 6.25 = 12.5$ The circle equation is $(x - 3.5)^2 + (y - 4.5)^2 = 12.5$, which is choice C. Choice A has center $(3, 4)$ and radius squared $13$ - this center doesn't match the diagonal midpoint. Choice B uses center $(4, 5)$ with radius squared $10$ - again, wrong center location. Choice D places the center at $(2, 3)$ with radius squared $18$ - this center is also incorrect and would place the circle too far from the actual vertices. Strategy tip: For any quadrilateral inscribed in a circle, the center is where the perpendicular bisectors of the sides meet. For rectangles specifically, this simplifies to the midpoint of either diagonal, making the calculation much faster.
4. The line $3x + 4y = 24$ intersects the coordinate axes to form a triangle with the origin. What is the equation of the circle inscribed in this triangle?
   Answer: $(x - 2)^2 + (y - 2)^2 = 4$ First, find the intercepts. For $x$-intercept: set $y = 0$, so $3x = 24$ and $x = 8$. For $y$-intercept: set $x = 0$, so $4y = 24$ and $y = 6$. The triangle has vertices at $(0, 0)$, $(8, 0)$, and $(0, 6)$. For a right triangle with legs of length $a$ and $b$ and hypotenuse $c$, the inradius is $r = \frac{a + b - c}{2}$. Here, $a = 8$, $b = 6$, and $c = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$. So $r = \frac{8 + 6 - 10}{2} = \frac{4}{2} = 2$. The incenter of a right triangle with legs along the axes is at $(r, r) = (2, 2)$. Therefore, the equation is $(x - 2)^2 + (y - 2)^2 = 4$. Choice B has the wrong center and radius. Choices C and D have incorrect centers that don't correspond to the incenter of this triangle.
5. A circle has its center at $(-2, 3)$ and passes through the point $(1, 7)$. If this circle intersects the $y$-axis at two points, what is the $y$-coordinate of the lower intersection point?
   Answer: $3 - \sqrt{21}$ When you encounter a circle problem involving intersection points, you need to find the circle's equation first, then solve for where it meets the given line or axis. Start by finding the radius using the distance formula between the center $(-2, 3)$ and the given point $(1, 7)$: $r = \sqrt{(1-(-2))^2 + (7-3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ The circle's equation is $(x + 2)^2 + (y - 3)^2 = 25$. To find where this circle intersects the $y$-axis, substitute $x = 0$: $(0 + 2)^2 + (y - 3)^2 = 25$ $4 + (y - 3)^2 = 25$ $(y - 3)^2 = 21$ $y - 3 = \pm\sqrt{21}$ $y = 3 \pm \sqrt{21}$ The two intersection points are at $y = 3 + \sqrt{21}$ (upper) and $y = 3 - \sqrt{21}$ (lower). The lower point is $3 - \sqrt{21}$, which is choice D. Choice A ($3 - 2\sqrt{6}$) likely comes from incorrectly calculating $2\sqrt{6} = \sqrt{24}$ instead of $\sqrt{21}$. Choice B ($3 - \sqrt{6}$) might result from forgetting to square the radius or making an arithmetic error. Choice C ($3 - \sqrt{15}$) could come from incorrectly computing $25 - 4 = 15$ but missing a step in the calculation. Remember: always verify your radius calculation first, then carefully substitute into the circle equation. Double-check your algebra when solving for intersection points.
6. Line $\ell$ is tangent to the circle $(x - 4)^2 + (y + 1)^2 = 25$ at the point $(8, 2)$. What is the equation of line $\ell$?
   Answer: $4x + 3y = 38$ A tangent line to a circle is perpendicular to the radius at the point of tangency. The center of the circle is (4, -1) and the point of tangency is (8, 2). The slope of the radius from (4, -1) to (8, 2) is (2-(-1))/(8-4) = 3/4. The tangent line has slope that is the negative reciprocal: -4/3. Using point-slope form with point (8, 2): y - 2 = -4/3(x - 8), which gives y - 2 = -4x/3 + 32/3, so y = -4x/3 + 32/3 + 2 = -4x/3 + 38/3. Multiplying by 3: 3y = -4x + 38, or 4x + 3y = 38.
7. In triangle $ABC$, the perpendicular bisector of side $AB$ passes through point $C(7, 3)$. If $A(-1, 5)$ and $B(3, -1)$, what is the equation of the line containing side $AC$?
   Answer: $y = -\frac{1}{4}x + \frac{19}{4}$ Since the perpendicular bisector of AB passes through C, triangle ABC is isosceles with CA = CB. The perpendicular bisector of AB has midpoint (1, 2) and slope 2/3 (negative reciprocal of AB's slope -3/2). Its equation is y - 2 = (2/3)(x - 1). We can verify C(7, 3) lies on this line: 3 - 2 = (2/3)(7 - 1) = 4, which is true. Now for line AC with A(-1, 5) and C(7, 3): slope = (3-5)/(7-(-1)) = -2/8 = -1/4. Using point-slope form with point A: y - 5 = (-1/4)(x - (-1)) = (-1/4)(x + 1), so y = -x/4 - 1/4 + 5 = -x/4 + 19/4.
8. A line is tangent to the circle $(x - 3)^2 + (y + 2)^2 = 25$ at the point $(7, 1)$. What is the equation of this tangent line?
   Answer: $4x + 3y = 31$ The tangent line at a point on a circle is perpendicular to the radius at that point. The center is $(3, -2)$ and the point of tangency is $(7, 1)$. The slope of the radius is $\frac{1-(-2)}{7-3} = \frac{3}{4}$. Therefore, the slope of the tangent line is $-\frac{4}{3}$. Using point-slope form: $y - 1 = -\frac{4}{3}(x - 7)$, which gives $y - 1 = -\frac{4}{3}x + \frac{28}{3}$. Rearranging: $y = -\frac{4}{3}x + \frac{28}{3} + 1 = -\frac{4}{3}x + \frac{31}{3}$. Multiplying by 3: $3y = -4x + 31$, so $4x + 3y = 31$. Choice B uses the slope of the radius instead of its negative reciprocal. Choice C incorrectly switches the coefficients. Choice D has the wrong constant term.
9. A circle is inscribed in a square with vertices at $(0, 0)$, $(6, 0)$, $(6, 6)$, and $(0, 6)$. What is the equation of this inscribed circle?
   Answer: $(x - 3)^2 + (y - 3)^2 = 9$ An inscribed circle touches all four sides of the square. The center of the inscribed circle is at the center of the square, which is the midpoint of the diagonal from (0,0) to (6,6), giving us (3,3). The radius of the inscribed circle is half the side length of the square. The side length is 6, so the radius is 3. Therefore, the equation is (x-3)² + (y-3)² = 3² = 9.

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