Answer with Step-by-step explanation:
We are given that equation of circle
[tex](x+1)^2+(y-9)^2=145=(\sqrt{145})^2[/tex]
Compare with equation of circle
[tex](x-h)^2+(y-k)^2=r^2[/tex]
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=[tex]\frac{1-9}{-10+1}=\frac{8}{9}[/tex]
By using slope formula
Slope:[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
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=[tex]-\frac{1}{\frac{8}{9}}=-\frac{9}{8}[/tex]
Now, the equation of tangent passing through the point(-10,1) with slope-9/8 is given by
[tex]y-1=-\frac{9}{8}(x+10)[/tex]
By using point-slope form
[tex]y-y_1=m(x-x_1)[/tex]
a.Kamal has made an error to find the slope of tangent .
Slope of tangent=[tex]-\frac{9}{8}[/tex]
But he used slope of tangent 8/9
b.The equation of tangent line
[tex]y-1=-\frac{9}{8}(x+10)[/tex]
