The equation of the tangent to the circle x^2 | vedclass

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(B) Let the equation of the tangent be $\frac{x}{x_1} + \frac{y}{y_1} = 1$. The area of the triangle formed by this line with the coordinate axes is $

Vedclass · Accepted Answer

$x \pm y = \pm a\sqrt{2}$

Vedclass · Answer

(B) Let the equation of the tangent be $\frac{x}{x_1} + \frac{y}{y_1} = 1$. The area of the triangle formed by this line with the coordinate axes is $\frac{1}{2} |x_1 y_1| = a^2$,so $|x_1 y_1| = 2a^2$. The line equation is $y_1 x + x_1 y - x_1 y_1 = 0$. Since this line is tangent to the circle $x^2 + y^2 = a^2$,the perpendicular distance from the origin $(0, 0)$ to the line must be equal to the radius $a$: $\frac{|-x_1 y_1|}{\sqrt{y_1^2 + x_1^2}} = a$ Squaring both sides: $\frac{x_1^2 y_1^2}{x_1^2 + y_1^2} = a^2$ Substitute $x_1^2 y_1^2 = (2a^2)^2 = 4a^4$: $\frac{4a^4}{x_1^2 + y_1^2} = a^2 \implies x_1^2 + y_1^2 = 4a^2$. We have $|x_1 y_1| = 2a^2$ and $x_1^2 + y_1^2 = 4a^2$. $(|x_1| + |y_1|)^2 = x_1^2 + y_1^2 + 2|x_1 y_1| = 4a^2 + 4a^2 = 8a^2 \implies |x_1| + |y_1| = 2a\sqrt{2}$. $(|x_1| - |y_1|)^2 = x_1^2 + y_1^2 - 2|x_1 y_1| = 4a^2 - 4a^2 = 0 \implies |x_1| = |y_1|$. Thus,$|x_1| = |y_1| = a\sqrt{2}$. Substituting these into the intercept form $\frac{x}{x_1} + \frac{y}{y_1} = 1$,we get $\frac{x}{\pm a\sqrt{2}} + \frac{y}{\pm a\sqrt{2}} = 1$,which simplifies to $x \pm y = \pm a\sqrt{2}$.

The equation of the tangent to the circle $x^2 + y^2 = a^2$ which makes a triangle of area $a^2$ with the coordinate axes is:

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