Question

7. Kamal gives the following proof that y−1=8/9(x+10) is the equation of a line that is tangent to a circle given by

Answer 1

Answer with Step-by-step explanation:

We are given that equation of circle

$(x+1)^2+(y-9)^2=145=(\sqrt{145})^2$

Compare with equation of circle

$(x-h)^2+(y-k)^2=r^2$

Where (h,k)=Center of circle

r=Radius of circle

We get circle of center=(-1,9)

Radius=12.04

The point (-10,1) is on the circle because it satisfied the circle equation .

Slope of radius=$\frac{1-9}{-10+1}=\frac{8}{9}$

By using slope formula

Slope:$m=\frac{y_2-y_1}{x_2-x_1}$

We know that radius is perpendicular to the tangent.

When two lines are perpendicular then their slope is opposite reciprocal to each other.

Slope of tangent=$-\frac{1}{\frac{8}{9}}=-\frac{9}{8}$

Now, the equation of tangent passing through the point(-10,1) with slope-9/8 is given by

$y-1=-\frac{9}{8}(x+10)$

By using point-slope form

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

a.Kamal has made an error to find the slope of  tangent .

Slope of tangent=$-\frac{9}{8}$

But he used slope of tangent 8/9

b.The equation  of tangent line

$y-1=-\frac{9}{8}(x+10)$