# Orthogonal Intersection of Two Circles

### Orthogonal Intersection of Two Circles:

#### 1. Orthogonal Intersection Definition:

• Two circles intersect orthogonally when their intersection points form right angles (90 degrees).

#### 2. Equation of Circles:

• Circles are defined by equations $\left(x-{h}_{1}{\right)}^{2}+\left(y-{k}_{1}{\right)}^{2}={r}_{1}^{2}$ and $\left(x-{h}_{2}{\right)}^{2}+\left(y-{k}_{2}{\right)}^{2}={r}_{2}^{2}$ with centers $\left({h}_{1},{k}_{1}\right)$ and $\left({h}_{2},{k}_{2}\right)$, and radii ${r}_{1}$ and ${r}_{2}$ respectively.

#### 3. Conditions for Orthogonal Intersection:

• Perpendicular Tangents: At the point of intersection, the tangents to each circle are perpendicular to each other.
• Product of Slopes: The product of the slopes of the tangents drawn from the point of intersection to each circle is equal to -1.

#### 4. Calculating Orthogonal Intersection:

• Step 1: Determine Tangents' Slopes:
• Calculate the slopes of the tangents by finding the derivatives of the circle equations at the points of intersection.
• Step 2: Check for Perpendicularity:
• Verify that the product of slopes is -1 to confirm orthogonal intersection.

#### 5. Example:

Consider the following circles:

${C}_{1}:\left(x-2{\right)}^{2}+\left(y-3{\right)}^{2}=4$ with center $\left(2,3\right)$ and radius ${r}_{1}=2$.

${C}_{2}:\left(x+1{\right)}^{2}+\left(y-1{\right)}^{2}=9$ with center $\left(-1,1\right)$ and radius ${r}_{2}=3$.

#### Steps to Determine Orthogonal Intersection:

Step 1: Find the Tangents' Slopes

Let's differentiate the circle equations to find the slopes of the tangents at the point of intersection.

For ${C}_{1}:\left(x-2{\right)}^{2}+\left(y-3{\right)}^{2}=4$:

• Differentiating implicitly, we get $\frac{dy}{dx}=-\frac{x-2}{y-3}$.
• At the point of intersection $\left({x}_{0},{y}_{0}\right)$, the slope of the tangent for ${C}_{1}$ will be $\frac{dy}{dx}$ evaluated at $\left({x}_{0},{y}_{0}\right)$.

For ${C}_{2}:\left(x+1{\right)}^{2}+\left(y-1{\right)}^{2}=9$:

• Similarly differentiating, we get $\frac{dy}{dx}=-\frac{x+1}{y-1}$.
• At the point of intersection $\left({x}_{0},{y}_{0}\right)$, the slope of the tangent for ${C}_{2}$ will be $\frac{dy}{dx}$ evaluated at $\left({x}_{0},{y}_{0}\right)$.

Step 2: Check for Perpendicularity

Evaluate the slopes at the point of intersection and confirm that the product of slopes is -1 to verify orthogonal intersection.

### Calculation:

Let's assume the circles intersect at point $\left({x}_{0},{y}_{0}\right)$.

For ${C}_{1}$ at $\left({x}_{0},{y}_{0}\right)$ (${C}_{1}:\left(x-2{\right)}^{2}+\left(y-3{\right)}^{2}=4$:

• Calculate $\frac{dy}{dx}$ at $\left({x}_{0},{y}_{0}\right)$: $\frac{dy}{dx}=-\frac{{x}_{0}-2}{{y}_{0}-3}$.

For ${C}_{2}$ at $\left({x}_{0},{y}_{0}\right)$ (${C}_{2}:\left(x+1{\right)}^{2}+\left(y-1{\right)}^{2}=9$:

• Calculate $\frac{dy}{dx}$ at $\left({x}_{0},{y}_{0}\right)$: $\frac{dy}{dx}=-\frac{{x}_{0}+1}{{y}_{0}-1}$.

Check if the product of slopes is -1: $\left(-\frac{{x}_{0}-2}{{y}_{0}-3}\right)×\left(-\frac{{x}_{0}+1}{{y}_{0}-1}\right)=1$.

### Conclusion:

If the product of slopes is indeed -1 for the given values of ${x}_{0}$ and ${y}_{0}$, it confirms that the tangents to the circles at their intersection point are perpendicular. Hence, the circles intersect orthogonally at that point.