Equation of the common tangent, with positive slope, to the circle x² + y² - 8x = 0 as well as to the hyperbola \(\rm \frac{x^2}{9}-\frac{y^2}{4}=1\), is:

Concept:

• The equation of a line, with slope m, is: y = mx + c.

• The distance between a point P(x₁, y₁) and the line ax + by + c = 0 is given by: Distance = \(\rm \frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\).

• The equation of a circle with center at O(a, b) and radius r, is given by: (x - a)² + (y - b)² = r².
• Tangent to a Hyperbola: If the line y = mx + c touches the hyperbola \(\rm \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), then c² = a²m² - b². The equation of the tangent is: \(\rm y = mx \pm \sqrt{a^2m^2 - b^2}\). Either of the lines is the equation of the tangent but not both.

Calculation:

The equation of the circle can be written as (x - 4)2 + y2 = 42.

Comparing with the general form of a circle, we have center O(4, 0) and radius r = 4.

The equation of the given hyperbola can be written as \(\rm \frac{x^2}{3^2}-\frac{y^2}{2^2}=1\).

Comparing with the general form of a hyperbola, we have a = 3 and b = 2.

The equation of the tangent to this hyperbola will have the form:

\(\rm y = mx \pm \sqrt{a^2m^2 - b^2}\)

⇒ \(\rm y = mx \pm \sqrt{9m^2 - 4}\)

Since this line is a tangent to the circle as well, we must have:

Distance from the center O(4, 0) of the circle to the tangent \(\rm y = mx \pm \sqrt{9m^2 - 4}\) = radius (r = 4) of the circle.

Using the formula for the distance of a point from a line, we get:

\(\rm \frac{\left|m(4)+(-1)(0)\pm\sqrt{9m^2-4}\right|}{\sqrt{m^2+(-1)^2}}=4\)

⇒ \(\rm \left|4m\pm\sqrt{9m^2-4}\right|=4\sqrt{m^2+1}\)

On squaring both sides, we get:

⇒ \(\rm 16m^2 \pm 8m\sqrt{9m^2-4}+9m^2-4=16m^2+16\)

⇒ \(\rm \pm 8m\sqrt{9m^2-4}=20-9m^2\)

Squaring again, we get:

⇒ \(\rm 64m^2(9m^2-4)=400+81m^4-360m^2\)

⇒ \(\rm 576m^4-256m^2=400+81m^4-360m^2\)

⇒ \(\rm 495 m^4 + 104 m^2 - 400=0\)

⇒ \(\rm m^2 = \frac{-104 \pm \sqrt{104^2-4(495)(-400)}}{2\times495}\)

⇒ \(\rm m^2 = \frac{-104 \pm \sqrt{802816}}{990}\)

⇒ \(\rm m^2 = \frac{-104 \pm896}{990}\)

Discarding the negative value of m²:

⇒ \(\rm m^2 = \frac{-104 +896}{990}=\frac45\)

Since the slope is given to be positive, we get:

⇒ \(\rm m=\frac{2}{\sqrt5}\)

∴ Equation of the tangent will be:

\(\rm y = mx \pm \sqrt{9m^2 - 4}\)

⇒ \(\rm y = \frac{2}{\sqrt5}x \pm \sqrt{9\left(\frac45\right) - 4}\)

⇒ \(\rm y = \frac{2}{\sqrt5}x \pm\frac{4}{\sqrt5}\)

⇒ \(\rm 2x-\sqrt5y\pm4=0\).

Additional Information

• The slope (m) of the tangent at a point P(a, b) to a curve y = f(x), is given by: \(\rm m=\left(\frac{dy}{dx}\right)_{x=a,y=b}\).
• Tangent to a Parabola: The equation of the tangent to the parabola y2 = 4ax, at a point (x1, y1), is given by: yy1 = 2a(x + x1).

  Normal to a Parabola: The equation of the normal to the parabola y2 = 4ax, at a point (x1, y1), is given by: 2a(y - y1) = (-y1)(x - x1).

• Tangent to a Circle: The equation of the tangent to the circle x² + y² = r² at a point (x₁, y₁), is given by: xx₁ + yy₁ = r².

  Normal to a Circle: The equation of a normal to the circle x² + y² = r² at a point (x₁, y₁), is given by: yx₁ - xy₁ = 0.

• Tangent to an Ellipse: The equation of the tangent to the ellipse \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), at a point (x₁, y₁), is given by: \(\rm \frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1\).

  Normal to an Ellipse: The equation of the normal to the ellipse \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), at a point (x₁, y₁), is given by: \(\rm \frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1\).

• Tangent to a Hyperbola: The equation of the tangent to the hyperbola \(\rm \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), at a point (x₁, y₁), is given by: \(\rm \frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1\).

  Normal to a Hyperbola: The equation of the normal to the hyperbola \(\rm \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), at a point (x₁, y₁), is given by: \(\rm \frac{a^2x}{x_1}+\frac{b^2y}{y_1}= a^2+b^2\).