Maths Orthogonal Circles
Orthogonal Circles
Orthogonal circles are a pair of circles that intersect at right angles. The point of intersection is called the radical center. The radical center is the same distance from any point on either circle.
Properties of Orthogonal Circles
 The radical center of two orthogonal circles is the midpoint of the line segment connecting their centers.
 The sum of the radii of two orthogonal circles is equal to the distance between their centers.
 The product of the radii of two orthogonal circles is equal to the square of the distance between their centers.
 The angle between the tangents to two orthogonal circles at their point of intersection is 90 degrees.
Applications of Orthogonal Circles
Orthogonal circles are used in a variety of applications, including:
 Gear design: Orthogonal circles are used to design gears that mesh smoothly.
 Bearing design: Orthogonal circles are used to design bearings that reduce friction.
 Cam design: Orthogonal circles are used to design cams that control the motion of other parts.
 Optics: Orthogonal circles are used to design lenses and mirrors that focus light.
Orthogonal circles are a versatile tool that can be used in a variety of applications. Their unique properties make them ideal for designing gears, bearings, cams, and optical components.
How to Draw a pair of Orthogonal Circles
Materials:
 A pair of compasses
 A ruler
 A pencil
 A piece of paper
Steps:

Draw a circle. Use the compasses to draw a circle of any size on the paper.

Draw a diameter. Use the ruler to draw a diameter of the circle. This is a straight line that passes through the center of the circle and intersects the circle at two points.

Draw a perpendicular diameter. Use the ruler to draw a perpendicular diameter of the circle. This is a straight line that intersects the first diameter at a right angle.

Draw the second circle. Use the compasses to draw a second circle of the same size as the first circle, centered on the intersection of the two diameters.

Erase the construction lines. Erase the diameter and perpendicular diameter that you drew in steps 2 and 3.
Tips:
 To make sure that the circles are perfectly orthogonal, use a protractor to measure the angle between the two diameters. It should be 90 degrees.
 You can also use a Tsquare or a right triangle to draw the perpendicular diameter.
 If you want to draw a pair of circles that are not the same size, simply use differentsized compasses to draw the circles.
Orthogonality Theorem
The Orthogonality Theorem states that if two vectors are orthogonal, then their dot product is zero. In other words, if $\mathbf{a}$ and $\mathbf{b}$ are orthogonal, then $\mathbf{a} \cdot \mathbf{b} = 0$.
Proof
The proof of the Orthogonality Theorem is simple. Let $\mathbf{a}$ and $\mathbf{b}$ be two vectors in an inner product space. Then, the dot product of $\mathbf{a}$ and $\mathbf{b}$ is defined as
$$\mathbf{a} \cdot \mathbf{b} = \sum_{i=1}^n a_ib_i$$
where $a_i$ and $b_i$ are the components of $\mathbf{a}$ and $\mathbf{b}$, respectively.
If $\mathbf{a}$ and $\mathbf{b}$ are orthogonal, then they are perpendicular to each other. This means that the angle between $\mathbf{a}$ and $\mathbf{b}$ is $90^\circ$.
The cosine of the angle between two vectors is defined as
$$\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{ \mathbf{a}  \mathbf{b} }$$
where $ \mathbf{a} $ and $ \mathbf{b} $ are the magnitudes of $\mathbf{a}$ and $\mathbf{b}$, respectively.
Since $\mathbf{a}$ and $\mathbf{b}$ are orthogonal, the angle between them is $90^\circ$. Therefore, $\cos\theta = 0$.
Substituting $\cos\theta = 0$ into the equation for the dot product, we get
$$\mathbf{a} \cdot \mathbf{b} =  \mathbf{a}   \mathbf{b}  \cos\theta = 0$$
Therefore, the dot product of two orthogonal vectors is zero.
The Orthogonality Theorem is a fundamental theorem of linear algebra. It has many applications in mathematics and physics.
Orthogonal Circles Solved Examples
Example 1: Finding the Radii of Orthogonal Circles
Problem: Two circles are orthogonal to each other. The radius of one circle is 3 cm. Find the radius of the other circle.
Solution:
Let the radius of the other circle be $r$. Since the circles are orthogonal, their radii are perpendicular to each other. Therefore, we can use the Pythagorean theorem to find $r$.
$$(3 \text{ cm})^2 + r^2 = (5 \text{ cm})^2$$
$$9 \text{ cm}^2 + r^2 = 25 \text{ cm}^2$$
$$r^2 = 16 \text{ cm}^2$$
$$r = 4 \text{ cm}$$
Therefore, the radius of the other circle is 4 cm.
Example 2: Finding the Centers of Orthogonal Circles
Problem: Two circles are orthogonal to each other. The center of one circle is at the point $(2, 3)$. The radius of the other circle is 5 cm. Find the center of the other circle.
Solution:
Let the center of the other circle be $(x, y)$. Since the circles are orthogonal, their radii are perpendicular to each other. Therefore, the line connecting the centers of the two circles is perpendicular to the line connecting the points $(2, 3)$ and $(x, y)$.
The slope of the line connecting the points $(2, 3)$ and $(x, y)$ is:
$$m = \frac{y  3}{x  2}$$
The slope of the line connecting the centers of the two circles is perpendicular to this, so its slope is:
$$m’ = \frac{1}{m} = \frac{x  2}{y  3}$$
We also know that the distance between the centers of the two circles is equal to the sum of their radii, which is 3 cm + 5 cm = 8 cm. Therefore, we can use the distance formula to find the coordinates of $(x, y)$.
$$d = \sqrt{(x  2)^2 + (y  3)^2} = 8 \text{ cm}$$
$$(x  2)^2 + (y  3)^2 = 64 \text{ cm}^2$$
Substituting the expression for $m’$ into this equation, we get:
$$(x  2)^2 + \left(\frac{x  2}{y  3}\right)^2 = 64 \text{ cm}^2$$
Simplifying this equation, we get:
$$(x  2)^2 + \frac{(x  2)^2}{(y  3)^2} = 64 \text{ cm}^2$$
$$(x  2)^2\left[1 + \frac{1}{(y  3)^2}\right] = 64 \text{ cm}^2$$
$$(x  2)^2\left[\frac{(y  3)^2 + 1}{(y  3)^2}\right] = 64 \text{ cm}^2$$
$$(x  2)^2\left[\frac{y^2  6y + 10}{(y  3)^2}\right] = 64 \text{ cm}^2$$
$$(x  2)^2 = \frac{64 \text{ cm}^2 (y  3)^2}{y^2  6y + 10}$$
$$x  2 = \pm \sqrt{\frac{64 \text{ cm}^2 (y  3)^2}{y^2  6y + 10}}$$
$$x = 2 \pm \sqrt{\frac{64 \text{ cm}^2 (y  3)^2}{y^2  6y + 10}}$$
Since the center of the circle must be a real number, we must choose the positive square root. Therefore,
$$x = 2 + \sqrt{\frac{64 \text{ cm}^2 (y  3)^2}{y^2  6y + 10}}$$
We can now substitute this expression for $x$ into the equation for $m’$ to find $y$.
$$m’ = \frac{x  2}{y  3} = \frac{2 + \sqrt{\frac{64 \text{ cm}^2 (y  3)^2}{y^2  6y + 10}}  2}{y  3}$$
$$m’ = \frac{\sqrt{\frac{64 \text{ cm}^2 (y  3)^2}{y^2  6y + 10}}}{y  3}$$
$$y  3 = \frac{y  3}{\sqrt{\frac{64 \text{ cm}^2 (y  3)^2}{y^2  6y + 10}}}$$
$$(y  3)^2 = \frac{(y  3)^2}{\frac{64 \text{ cm}^2 (y  3)^2}{y^2  6y + 10}}$$
$$(y  3)^2\left[1 + \frac{1}{\frac{64 \text{ cm}^2 (y  3)^2}{y^2  6y + 10}}\right] = 0$$
$$(y  3)^2\left[\frac{y^2  6y + 10 + 64 \text{ cm}^2}{y^2  6y + 10}\right] = 0$$
$$(y  3)^2\left[\frac{y^2 + 58y + 64 \text{ cm}^2}{y^2  6y + 10}\right] = 0$$
Since the square of a real number cannot be negative, we must have:
$$y^2 + 58y + 64 \text{ cm}^2 = 0$$
This is a quadratic equation, which we can solve using the quadratic formula:
$$y = \frac{b \pm \sqrt{b^2  4ac}}{2a}$$
In this case, $a = 1$, $b = 58$, and $c = 64 \text{ cm}^2$. Substituting these values into the quadratic formula, we get:
$$y = \frac{58 \pm \sqrt{58^2  4(1)(64 \text{ cm}^2)}}{2(1)}$$
$$y = \frac{58 \pm \sqrt{3364  256 \text{ cm}^2}}{2}$$
$$y = \frac{58 \pm \sqrt{3108 \text{ cm}^2}}{2}$$
$$y = \frac{58 \pm 56 \text{ cm}}{2}$$
Therefore, the two possible values for $y$ are:
$$y_1 = \frac{58 + 56 \text{ cm}}{2} = 1 \text{ cm}$$
$$y_2 = \frac{58  56 \text{ cm}}{2} = 57 \text{ cm}$$
Therefore, the two possible centers of the other circle are $(1, 57)$ and $(1, 1)$.
Orthogonal Circles FAQs
What are orthogonal circles?
Orthogonal circles are two circles that intersect at right angles. The point of intersection is called the radical center.
What are the properties of orthogonal circles?
The following are some of the properties of orthogonal circles:
 The radical center of two orthogonal circles is the midpoint of the line segment connecting their centers.
 The sum of the radii of two orthogonal circles is equal to the distance between their centers.
 The product of the radii of two orthogonal circles is equal to the square of the distance between their centers.
 The angle between the tangents to two orthogonal circles at their point of intersection is 90 degrees.
How can you find the radical center of two orthogonal circles?
There are a few different ways to find the radical center of two orthogonal circles. One way is to use the following formula:
$$Radical\ center = ((x1 + x2) / 2, (y1 + y2) / 2)$$
where (x1, y1) and (x2, y2) are the coordinates of the centers of the two circles.
Another way to find the radical center of two orthogonal circles is to construct the perpendicular bisectors of the line segment connecting their centers. The point of intersection of the perpendicular bisectors is the radical center.
What are some applications of orthogonal circles?
Orthogonal circles have a number of applications in geometry, including:
 Finding the circumcenter of a triangle
 Finding the incenter of a triangle
 Finding the excenters of a triangle
 Constructing regular polygons
 Solving geometric problems involving circles
Conclusion
Orthogonal circles are a useful tool in geometry. They have a number of properties that can be used to solve a variety of geometric problems.