Question

Consider the circle of radius 10 centered at the origin. provide answers accurate to two decimal places. (a) the equation of the tangent line to the circle through the point (-6,8) has equation y= x + . (b) suppose that l is a tangent line to this circle which is parallel to the line y=5x+7 and has a negative y intercept. then the point of tangency of l with this circle is ( , ).

Answer 1

(a) The equation of the tangent line can be written as
.. 3(x +6) -4(y -8) = 0
.. 3x -4y = -50
.. y = (3/4)x +25/2 . . . the equation of the tangent line


(b) The point of tangency will be the intersection of the circle with the perpendicular line through the circle center, y = (-1/5)x. A graphing calculator shows that point to be
.. (9.81, -1.96)

Answer 2

Answer:

(a) $y=\frac{3}{4}x+\frac{25}{2}$.

(b) $(\frac{25\sqrt{26}}{13}, -\frac{5\sqrt{26}}{13})$.

Step-by-step explanation:

(a)

The center of the circle is (0,0) and radius of the circle is 10 units.

The equation of circle is

$(x-0)^2+(y-0)^2=(10)^2$

$x^2+y^2=100$

We need to find the equation of the tangent line to the circle through the point (-6,8).

The radius of the circle is perpendicular to the tangent at point of tangency. So, slope of perpendicular line is

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{8-0}{-6-0}=-\frac{4}{3}$

Product of slopes of two perpendicular line is -1. So, the slope of tangent is 3/4.

Point slope form of line is

$y-y_1=m(x-x_1)$

The slope of tangent is 3/4 and it passes through the point (-6,8).

$y-8=\frac{3}{4}(x-(-6))$

$y-8=\frac{3}{4}(x}+\frac{9}{2}$

$y=\frac{3}{4}(x}+\frac{9}{2}+8$

$y=\frac{3}{4}(x}+\frac{25}{2}$

Therefore, the equation of tangent is $y=\frac{3}{4}x+\frac{25}{2}$.

(b)

We need to find the tangent line to this circle which is parallel to the line y=5x+7 and has a negative y intercept.

The slope of two parallel lines are same. So, the slope of required tangent is 5.

The slope of radius which connects (0,0) and the point of intersection is -1/5 because it is perpendicular to the line whose slope is 5. So, the equation of perpendicular line is

$y=-\frac{1}{5}x$

Substitute $y=-\frac{1}{5}x$ in the equation of circle.

$x^2+(-\frac{1}{5}x)^2=100$

$x^2+\frac{1}{25}x^2=100$

$\frac{26}{25}x^2=100$

$x^2=\frac{2500}{26}$

$x^2=\frac{1250}{13}$

Taking square root on both sides.

$x=\pm \sqrt{\frac{1250}{13}}=\frac{25\sqrt{26}}{13}$

The value of x is $\frac{25\sqrt{26}}{13}$ and $-\frac{25\sqrt{26}}{13}$.

The value of x is

$y=-\frac{1}{5}x=-\frac{1}{5}\times \frac{25\sqrt{26}}{13}=-\frac{5\sqrt{26}}{13}$

$y=-\frac{1}{5}x=-\frac{1}{5}\times (-\frac{25\sqrt{26}}{13})=\frac{5\sqrt{26}}{13}$

We need negative y intercept.

Therefore, the point of tangency of l with this circle is $(\frac{25\sqrt{26}}{13}, -\frac{5\sqrt{26}}{13})$.