Question Statement

Find the equation of the common tangent to the circles: $x^2+y^2-2y-3=0$ $x^2+y^2-8x+4y+11=0$ given that they touch each other externally.

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Background and Explanation

To solve this problem, we need to identify the centers and radii of both circles. When two circles touch externally, their point of contact divides the line segment joining their centers internally in the ratio of their radii. We can then find the slope of the tangent at that point using implicit differentiation and determine the line's equation.

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Solution

The equations of the given circles are:

1. $x^2+y^2-2y-3=0\dots \text{(i)}$
2. $x^2+y^2-8x+4y+11=0\dots \text{(ii)}$

1. Find the Centers and Radii

For circle (i), the center $C_1$ is found by taking half the coefficients of $x$ and $y$ and changing the sign: $C_1\left(\frac{0}{-2},\frac{-2}{-2}\right)=C_1(0,1)$ The radius $r_1$ is: $r_1=\sqrt{0^2+1^2-(-3)}=\sqrt{1+3}=2$

For circle (ii), the center $C_2$ is: $C_2\left(\frac{-8}{-2},\frac{4}{-2}\right)=C_2(4,-2)$ The radius $r_2$ is: $r_2=\sqrt{4^2+(-2{)}^2-11}=\sqrt{16+4-11}=\sqrt{9}=3$

2. Find the Point of Contact

Since the circles touch externally, the common point $P$ divides the line segment $C_1{C}_2$ internally in the ratio $r_1:r_2$, which is $2:3$. Using the ratio formula: $P=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$ $P=\left(\frac{2(4)+3(0)}{2+3},\frac{2(-2)+3(1)}{2+3}\right)$ $P=\left(\frac{8}{5},-\frac{1}{5}\right)$

3. Find the Slope of the Tangent

To find the slope of the tangent line, we differentiate the equation of circle (i) with respect to $x$: $2x+2y\frac{dy}{dx}-2\frac{dy}{dx}+0=0$ Divide by 2: $x+(y-1)\frac{dy}{dx}=0$ $\frac{dy}{dx}=\frac{-x}{y-1}$

Now, evaluate the slope at point $P\left(\frac{8}{5},-\frac{1}{5}\right)$: $\frac{dy}{dx}=\frac{-\frac{8}{5}}{-\frac{1}{5}-1}=\frac{-\frac{8}{5}}{-\frac{6}{5}}=\frac{8}{6}=\frac{4}{3}$ Thus, the slope $m$ of the tangent line is $\frac{4}{3}$.

4. Determine the Equation of the Tangent

Using the point-slope form $y-y_1=m(x-x_1)$: $y-\left(-\frac{1}{5}\right)=\frac{4}{3}\left(x-\frac{8}{5}\right)$ $y+\frac{1}{5}=\frac{4}{3}x-\frac{32}{15}$

Multiply the entire equation by 15 to clear the denominators: $15y+3=20x-32$ Rearranging the terms into general form: $20x-15y-35=0$ Divide the entire equation by 5: $4x-3y-7=0$

The required equation of the common tangent line is $4x-3y-7=0$.

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Key Formulas or Methods Used

• Circle Center and Radius: For $x^2+y^2+2gx+2fy+c=0$, Center is $(-g,-f)$ and $r=\sqrt{g^2+f^2-c}$.
• Section Formula (Internal Division): $P=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$.
• Implicit Differentiation: Used to find the derivative $\frac{dy}{dx}$ which represents the slope of the tangent.
• Point-Slope Form of a Line: $y-y_1=m(x-x_1)$.

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Summary of Steps

1. Calculate the center and radius for both given circles.
2. Use the ratio of the radii ($2:3$) and the section formula to find the coordinates of the point of contact $P$.
3. Differentiate one of the circle equations to find the expression for the slope $\frac{dy}{dx}$.
4. Substitute the coordinates of $P$ into the derivative to find the specific slope $m$.
5. Apply the point-slope formula to find the final linear equation of the tangent.