$Finding;the;Tangent;to;a;Unit;Circle$

The problem is to find the equation of the tangent to a unit circle at a given point.

Given a unit circle with the equation:

$x^2 + y^2 = 1$

and a point \((a, b)\) on the circle. Since \((a, b)\) is on the circle, we have:

$a^2 + b^2 = 1$

To find the tangent line at \((a, b)\), we use the fact that the radius of the circle at this point is perpendicular to the tangent. The slope of the radius is:

$m_r = \frac{b}{a}$

Thus, the slope of the tangent line, being the negative reciprocal, is:

$m_t = -\frac{a}{b}$

Using the point-slope form of the line equation:

$y – b = m_t (x – a)$

Substitute the values:

$y – b = -\frac{a}{b}(x – a)$

Simplify to get the equation of the tangent line in slope-intercept form:

$bx + ay = 1$

Given the unit circle:

$x^2 + y^2 = 1$

and a point of tangency ((a, b)), where:

$a^2 + b^2 = 1$

We seek the tangent line at ((a, b)). The derivative of the circle equation implicitly is:

$2x + 2yfrac{dy}{dx} = 0$

Solving for (frac{dy}{dx}) gives:

$frac{dy}{dx} = -frac{x}{y}$

At the point ((a, b)), the slope of the tangent line is:

$m_t = -frac{a}{b}$

Using the point-slope form:

$y – b = m_t (x – a)$

Substitute (m_t) and ((a, b)):

$y – b = -frac{a}{b}(x – a)$

Multiply through by (b) to clear the fraction:

$by – b^2 = -a(x – a)$

Rearrange to standard form:

$bx + ay = 1$

Given the circle:

$x^2 + y^2 = 1$

and point ((a, b)) on it, so:

$a^2 + b^2 = 1$

The slope is:

$m_t = -frac{a}{b}$

The tangent line is:

$y – b = -frac{a}{b}(x – a)$

Simplify:

$bx + ay = 1$

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