# 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 $(x-{h}_{1}{)}^{2}+(y-{k}_{1}{)}^{2}={r}_{1}^{2}$ and $(x-{h}_{2}{)}^{2}+(y-{k}_{2}{)}^{2}={r}_{2}^{2}$ with centers $({h}_{1},{k}_{1})$ and $({h}_{2},{k}_{2})$, 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:

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

${C}_{2}:(x+1{)}^{2}+(y-1{)}^{2}=9$ with center $(-1,1)$ 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 ${C}_{1}:(x-2{)}^{2}+(y-3{)}^{2}=4$:

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

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

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

**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 $({x}_{0},{y}_{0})$.

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

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

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

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

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

### 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.