Question

The circle C has centre (-1; 5) and radius 5.

i) Show that the equation of C can be written as x2 + y2 + 2x - 10y + 1 = 0.

ii) Find the equation of the tangent to C at the point A ≡ (-4; 9).

The point B lies on the circle C such that AB is a diameter.

(iii) Find the equation of the circle through the points A; B and O, where O is the origin.

Answer 1

i)

$$(x-(-1))^2+(y-5)^2=(5)^2$$

$$x^2+2x+1+y^2-10y+25=25$$

$$x^2+y^2+2x-10y+1=0$$

ii)

$$x^2+y^2+2x-10y+1=0$$

Differentiate both sides with respect to $x$ and use the Chain Rule

$$2x+2yy'+2-10y'=0$$

$$y'=\dfrac{x+1}{5-y}$$

Point $A = (-4, 9).$

$$slope=m=y'(-4)=\dfrac{-4+1}{5-9}=\dfrac{3}{4}$$

The equation of the tangent to C at the point $A = (-4, 9)$ in point-slope form

$$y-9=\dfrac{3}{4}(x-(-4))$$

The equation of the tangent to C at the point $A = (-4, 9)$ in slope-intercept form

$$y=\dfrac{3}{4}x+12$$

iii) Point $C(-1,5)$ is the middle of $AB$

$$-1=\dfrac{-4+x_B}{2}=>x_B=2$$

$$5=\dfrac{9+y_B}{2}=>y_B=1$$

Point $B(2,1).$

$$(0-m)^2+(0-n)^2=r^2$$ $$(-4-m)^2+(9-n)^2=r^2$$ $$(2-m)^2+(1-n)^2=r^2$$

$$m^2+n^2=r^2$$ $$16+8m+81-18n=0$$ $$4-4m+1-2n=0$$

$$m^2+n^2=r^2$$ $$107-22n=0$$ $$m=-\dfrac{1}{2}n+\dfrac{5}{4}$$

$$m=-\dfrac{13}{11}, n=\dfrac{107}{22}$$

$$r^2=(-\dfrac{13}{11})^2+(\dfrac{107}{22})^2=\dfrac{12125}{484}$$

The equation of the circle

$$(x+\dfrac{13}{11})^2+(y-\dfrac{107}{22})^2=\dfrac{12125}{484}$$