Question Statement

$A(-5,-1)$ and $B(1,5)$ are two points on the circle $x^2+y^2=26$.

1. Find the point of intersection of the tangents to the circle at $A$ and $B$.
2. Show that the point of intersection of the tangent lines, the midpoint of chord $\overline{AB}$, and the centre of the circle are collinear.

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Background and Explanation

To solve this problem, we use implicit differentiation to find the slope of the tangent line at any point on a circle. We then use the point-slope form to find the equations of the tangents and solve them as a system of linear equations to find their intersection. Finally, we use the slope formula to verify collinearity.

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Solution

Part (a): Finding the Point of Intersection

The equation of the circle is: $x^2+y^2=26$

To find the slope of the tangent line at any point $(x,y)$, we differentiate the equation with respect to $x$: $2x+2y\frac{dy}{dx}=0$ $\Rightarrow \frac{dy}{dx}=-\frac{2x}{2y}=-\frac{x}{y}$

Tangent at point $A(-5,-1)$: The slope $m_1$ at $A$ is: $\frac{dy}{dx}=-\frac{-5}{-1}=-5$ Using the point-slope form $y-y_1=m(x-x_1)$: $y-(-1)=-5(x-(-5))$ $y+1=-5(x+5)$ $y+1=-5x-25$ $5x+y=-26\dots \text{(i)}$

Tangent at point $B(1,5)$: The slope $m_2$ at $B$ is: $\frac{dy}{dx}=-\frac{1}{5}$ Using the point-slope form: $y-5=-\frac{1}{5}(x-1)$ $5(y-5)=-1(x-1)$ $5y-25=-x+1$ $x+5y=26\dots \text{(ii)}$

Finding the Intersection Point $I$: We solve equations (i) and (ii) simultaneously. Multiply equation (ii) by 5: $5(x+5y)=5(26)\Rightarrow 5x+25y=130\dots \text{(iii)}$ Subtract equation (i) from equation (iii): $(5x+25y)-(5x+y)=130-(-26)$ $24y=156$ $y=\frac{156}{24}=\frac{13}{2}$

Substitute $y=\frac{13}{2}$ into equation (i): $5x+\frac{13}{2}=-26$ $5x=-26-\frac{13}{2}$ $5x=\frac{-52-13}{2}=-\frac{65}{2}$ $x=-\frac{65}{10}=-\frac{13}{2}$

Thus, the point of intersection is $I\left(-\frac{13}{2},\frac{13}{2}\right)$.

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Part (b): Proving Collinearity

1. Find the Midpoint $M$ of chord $\overline{AB}$: Using the midpoint formula $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ for $A(-5,-1)$ and $B(1,5)$: $M=\left(\frac{-5+1}{2},\frac{-1+5}{2}\right)=\left(\frac{-4}{2},\frac{4}{2}\right)=(-2,2)$

2. Identify the Centre of the Circle $O$: For the circle $x^2+y^2=26$, the centre is $O(0,0)$.

3. Check Slopes for Collinearity: We check the slopes of the segments formed by $O(0,0)$, $M(-2,2)$, and $I\left(-\frac{13}{2},\frac{13}{2}\right)$. $\text{Slope of }\overline{OM}=\frac{2-0}{-2-0}=\frac{2}{-2}=-1$ $\text{Slope of }\overline{OI}=\frac{\frac{13}{2}-0}{-\frac{13}{2}-0}=\frac{\frac{13}{2}}{-\frac{13}{2}}=-1$

Since the slope of $\overline{OM}$ is equal to the slope of $\overline{OI}$ and they share a common point $O$, the points $O$, $M$, and $I$ are collinear.

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Key Formulas or Methods Used

• Implicit Differentiation: Used to find the slope of the tangent to the circle.
• Point-Slope Form: $y-y_1=m(x-x_1)$
• System of Linear Equations: Solved via the elimination method to find the intersection point.
• Midpoint Formula: $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$
• Slope Formula: $m=\frac{y_2-y_1}{x_2-x_1}$
• Collinearity Condition: Three points are collinear if the slopes between any two pairs are equal.

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Summary of Steps

1. Differentiate the circle equation $x^2+y^2=26$ to get the slope formula $\frac{dy}{dx}=-\frac{x}{y}$.
2. Calculate the specific slopes and equations for the tangents at points $A$ and $B$.
3. Solve the resulting linear equations to find the intersection point $I\left(-\frac{13}{2},\frac{13}{2}\right)$.
4. Calculate the midpoint $M$ of the segment $AB$.
5. Determine the slopes of $OM$ and $OI$ relative to the origin $(0,0)$.
6. Conclude collinearity because the slopes are identical.