Question Statement

Find the equation of the normal to the circle $x^2+y^2-2x+2y-11=0$ at the point on the circle with ordinate $-4$.

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Background and Explanation

To find the equation of a normal line to a circle, we first need to identify the specific points on the circle's circumference using the given coordinate (the ordinate, or $y$-value). A normal line is perpendicular to the tangent line at the point of contact; its slope is the negative reciprocal of the tangent's slope.

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Solution

The given equation of the circle is: $x^2+y^2-2x+2y-11=0\dots \text{(i)}$

Step 1: Find the points on the circle

We are given that the ordinate is $-4$, which means $y=-4$. To find the corresponding $x$-coordinates, we substitute $y=-4$ into equation (i):

$x^2+(-4{)}^2-2x+2(-4)-11=0$ $x^2+16-2x-8-11=0$ $x^2-2x-3=0$

Now, we solve this quadratic equation by factoring: $x^2-3x+x-3=0$ $x(x-3)+1(x-3)=0$ $(x+1)(x-3)=0$

This gives us two possible values for $x$: $x+1=0\Rightarrow x=-1$ $x-3=0\Rightarrow x=3$

Thus, there are two points on the circle where the ordinate is $-4$: $A(-1,-4)\text{and}B(3,-4)$

Step 2: Find the slope of the normal line

To find the slope of the tangent, we differentiate equation (i) implicitly with respect to $x$: $\frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)-\frac{d}{dx}(2x)+\frac{d}{dx}(2y)-\frac{d}{dx}(11)=0$ $2x+2y\frac{dy}{dx}-2+2\frac{dy}{dx}=0$

Divide the entire equation by $2$: $x+y\frac{dy}{dx}-1+\frac{dy}{dx}=0$

Rearrange to solve for the slope of the tangent ($\frac{dy}{dx}$): $(y+1)\frac{dy}{dx}=1-x$ $\frac{dy}{dx}=\frac{1-x}{y+1}$

The slope of the normal line ($m_n$) is the negative reciprocal of the tangent slope: $m_n=-\frac{1}{\frac{dy}{dx}}=-\frac{y+1}{1-x}=\frac{y+1}{x-1}$

Step 3: Equation of the normal at point $A(-1,-4)$

First, calculate the slope of the normal at $A$: $m_n=\frac{-4+1}{-1-1}=\frac{-3}{-2}=\frac{3}{2}$

Using the point-slope form $y-y_1=m(x-x_1)$: $y-(-4)=\frac{3}{2}(x-(-1))$ $y+4=\frac{3}{2}(x+1)$ $2(y+4)=3(x+1)$ $2y+8=3x+3$ $3x-2y-5=0$

Step 4: Equation of the normal at point $B(3,-4)$

First, calculate the slope of the normal at $B$: $m_n=\frac{-4+1}{3-1}=\frac{-3}{2}$

Using the point-slope form $y-y_1=m(x-x_1)$: $y-(-4)=-\frac{3}{2}(x-3)$ $y+4=-\frac{3}{2}(x-3)$ $2(y+4)=-3(x-3)$ $2y+8=-3x+9$ $3x+2y-1=0$

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Key Formulas or Methods Used

• Ordinate: Refers to the $y$-coordinate of a point.
• Implicit Differentiation: Used to find the derivative $\frac{dy}{dx}$ of the circle.
• Normal Slope: $m_{normal}=-\frac{1}{m_{tangent}}$.
• Point-Slope Form: $y-y_1=m(x-x_1)$.
• Quadratic Factoring: Used to find the $x$-intercepts/points.

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Summary of Steps

1. Substitute $y=-4$ into the circle equation to find the $x$-coordinates of the points of contact.
2. Differentiate the circle equation implicitly to find the expression for the slope of the tangent.
3. Determine the general expression for the slope of the normal line by taking the negative reciprocal of the tangent slope.
4. For each point found in Step 1, calculate the specific slope and use the point-slope formula to find the final linear equations.