To find the equation of the tangent line to the circle [tex]\( x^2 + y^2 = 1 \)[/tex] at the point [tex]\( Q \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \)[/tex], we use the following method.
First, recall that the equation of the tangent to a circle [tex]\( x^2 + y^2 = r^2 \)[/tex] at a point [tex]\( (x_1, y_1) \)[/tex] on the circle is given by:
[tex]\[ x_1 x + y_1 y = r^2 \][/tex]
For our specific case, the circle has the equation [tex]\( x^2 + y^2 = 1 \)[/tex], meaning [tex]\( r^2 = 1 \)[/tex]. Thus, the tangent line at point [tex]\( Q \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \)[/tex] can be found using:
[tex]\[ \frac{1}{2} x + \frac{\sqrt{3}}{2} y = 1 \][/tex]
Next, we need to rearrange this equation into the slope-intercept form [tex]\( y = ax + b \)[/tex].
Starting from the tangent equation:
[tex]\[ \frac{1}{2} x + \frac{\sqrt{3}}{2} y = 1 \][/tex]
Isolate the [tex]\( y \)[/tex]-term:
[tex]\[ \frac{\sqrt{3}}{2} y = 1 - \frac{1}{2} x \][/tex]
Multiply both sides by [tex]\( \frac{2}{\sqrt{3}} \)[/tex] to solve for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}} x \][/tex]
This simplifies to:
[tex]\[ y = \left( -\frac{1}{\sqrt{3}} \right) x + \frac{2}{\sqrt{3}} \][/tex]
Therefore, the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] in the equation [tex]\( y = ax + b \)[/tex] are:
[tex]\[ a = -\frac{1}{\sqrt{3}} \approx -0.5773502691896258 \][/tex]
[tex]\[ b = \frac{2}{\sqrt{3}} \approx 1.1547005383792517 \][/tex]
Thus, the equation of the tangent line is:
[tex]\[ y = -0.5773502691896258 x + 1.1547005383792517 \][/tex]
So, the value of [tex]\( a \)[/tex] is [tex]\( -0.5773502691896258 \)[/tex] and the value of [tex]\( b \)[/tex] is [tex]\( 1.1547005383792517 \)[/tex].
