Question

The equation of a circle is given as
X²+y²-8x+2y+13=0
What is the centre and radius of the circle
Prove that the point D(6,-1)lies on the circle
Find the equation of the tangent to the circle at that point

Answer 1

Let us rewrite the equation of circle in form of

$x^2-8x+y^2+2y+13=0, \\ (x-4)^2-16 + (y+1)^2 - 1 +13 = 0, \\ (x-4)^2 + (y+1)^2 = 2^2.$

Therefore, the center is located at (4,-1) and the radius is 2.

Let us substitute the coordinates of D in the equation obtained above.

$(6-4)^2 + (-1+1)^2 = 4, \\ 2^2 + 0^2 = 4.$

The equality is true, so D lies on the circle.

The equation of the tangent line to the circumference at point (x₁,y₁) with center at (a,b) and radius r

is $(x_1-a)(x-a) + (y_1-b)(y-b) = r^2.$

In our case

$(6-4)(x-4) + (-1-(-1))(y-(-1)) = 2^2, \;\;\; 2(x-4) = 4, \;\;\; x - 4 = 2, \;\;\; x = 6.$