Answer to Question #129230 in Analytic Geometry for Sym

Question #129230
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
1
Expert's answer
2020-08-12T17:41:43-0400

Let us rewrite the equation of circle in form of

x28x+y2+2y+13=0,(x4)216+(y+1)21+13=0,(x4)2+(y+1)2=22.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.

(64)2+(1+1)2=4,22+02=4.(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 (x1a)(xa)+(y1b)(yb)=r2.(x_1-a)(x-a) + (y_1-b)(y-b) = r^2.

In our case

(64)(x4)+(1(1))(y(1))=22,      2(x4)=4,      x4=2,      x=6.(6-4)(x-4) + (-1-(-1))(y-(-1)) = 2^2, \;\;\; 2(x-4) = 4, \;\;\; x - 4 = 2, \;\;\; x = 6.


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