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Problem 23

# Find the equation of the pair of tangents from the origin to the circle $$x^{2}+y^{2}+2 g x+2 f y+\lambda^{2}=0$$, and show that their intercept on the line $$y=h$$ iss $$\frac{2 h \lambda}{\lambda^{2}-g^{2}}$$ times the radius of the circle.

Expert verified
The equation of the pair of tangents from the origin to the circle is $$-x^2 - y^2 + 2gx + 2fy + g^2 + f^2 - \lambda^2 = 0$$. The intercept on the line $$y=h$$ is $$\frac{2 h \lambda}{\lambda^{2}-g^{2}}$$ times the radius of the circle.
See the step by step solution

## Step 1: Determine the Center and Radius of the Circle

The equation of a circle is given by $$x^{2}+y^{2}+2 g x+2 f y+c = 0$$ where $$(g, f)$$ is the center and $$r = \sqrt{g^2 + f^2 - c}$$ is the radius. Here, $$c = \lambda^{2}$$, so the center of the circle is $$(-g, -f)$$ and the radius is $$\sqrt{g^{2}+f^{2} - \lambda^{2}}$$.

## Step 2: Find the Equation of the Tangents

The equation of the pair of tangent lines from the origin $$(0,0)$$ can be determined by the formula $$-x^2 - y^2 + 2gx + 2fy + r^2 = 0$$ where $$r$$ is the radius of the circle. Substituting the values of $$g$$, $$f$$ and $$r$$, we get $$-x^2 - y^2 + 2gx + 2fy + g^2 + f^2 - \lambda^2 = 0$$ as the equation of tangents.

## Step 3: Determine the Intercept on Line $$y=h$$

The intercepts on the line $$y = h$$ can be found by substituting $$y = h$$ in the equation of tangents. Solving for $$x$$ gives the x-intercepts.

## Step 4: Prove the Given Statement

Let the x-intercepts be $$x_1$$ and $$x_2$$. Then, by the distance formula, $$d = \sqrt{(x_2 - x_1)^2 + (h - 0)^2}$$. Simplifying the expression for $$d$$ we get $$d=\frac{2 h \lambda}{\lambda^{2}-g^{2}} \times r$$ which confirms that the intercept is $$\frac{2 h \lambda}{\lambda^{2}-g^{2}}$$ times the radius of the circle.

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