Question

How can you find the equations of the tangents from the origin to the circle x^2 + y^2 - 10x - 6y + 25 = 0?

Answer 1

Please see the attachment

Answer 2

To find the equation of the tangent from the origin to the circle  $x^2+y^2-10x-6y+25=0$

Taking perfect square $x^2+y^2-10x-6y+25=0$

                     $(x-5)^2-25+(y-3)^2-9+25=0\\(x-5)^2+(y-3)^2=9$

                       

The equation of the circle is  $(x-5)^2+(y-3)^2=9$   (1)

Differentiating equation (1) with respect to x

$2(x-5)+\frac{2(y-3)dy}{dx}=0$  

$\frac{dy}{dx}=-\frac{x-5}{y-3}$  

Let (a,b) be any point on the circle which is tangent and the line through (a,b) with slope equal to the value of  $\frac{dy}{dx}$ at (a,b)

Substituting the point a,b in $\frac{dy}{dx}$

$\frac{dy}{dx}=-\frac{a-5}{b-3}$  

In the point slope form the equation the line through (a,b) is on the circle  

$y-b=-\frac{a-5}{(b-3)(x-a)}$  

Substituting (0,0) point on the above equation

$0-b=-\frac{0-5}{(b-3)(0-a)}\\ -b(b-3)=a(a-5)\\ a^2-5a+b^2-3b=0$

Substituting (a,b) in the given circle  

$a^2+b^2-10a-6b+25=0$  

The below two equations are enough to solve the two variables  

$a^2-5a+b^2-3b=0\\ a^2+b^2-10a-6b+25=0$

Subtract the above equation  

5a+3b-25=0  

$b=\frac{25-5a}{3}$  

Substituting the value of b in equation $a^2-5a+b^2-3b=0$

Then the equation becomes $34a^2-250a+400=0$

∴ The equation of the circle meets at (5, 0)

a=5 is one of the solution  

a=4017  

∴The value of a,b is (5,0) or (40/17,75/17)

$y-b=-\frac{a-5}{(b-3)(x-a)}$  

$y -\frac{75}{17} = -\frac{\frac{40}{17} - 5}{\frac{75}{17} - 3x -\frac{40}{17}}$

$y -\frac{75}{17} = -\frac{45}{24}(x - \frac{40}{17})\\y = \frac{45}{24} x$

∴ The equation of tangent from the origin to the circumference of the circle is $y = \frac{45}{24} x$