Question

The coordinates of the points A and B in the x-y plane are (-3,-4) and (5,0) respectively.
(a) Obtain the equation of the circle on AB as diameter, and show that the point (-3,0) lies on the circle.
(b) Find also:
(i) the equation of a line CD in parallel to the line 2x +2y-7=0 whose intercept on the x-axis is -3
(ii) the coordinates of the points of intersection of the line CD with the circle
(iii) the equation of the tangent of the circle at the point (-3,0)

Answer 1

(a) The center of a circle is the midpoint of a diameter:

$$x=\frac{x_A+x_B}{2}=\frac{-3+5}{2}=\frac{2}{2}=1$$

$$y=\frac{y_A+y_B}{2}=\frac{-4+0}{2}=\frac{-4}{2}=-2$$

The coordinates of the centre is (1,-2).

Since the point B lies on the circle, the coordinates (5,0) must satisfy the equation of the circle: $(x-h)^2+(y-k)^2 =r^2$, where (h, k) is the center of the circle and r is the radius.

$$(5-1)^2 +(0+2)^2 =r^2$$ $$16+4=r^2$$ $$r^2=20$$

The equation of the circle is: $(x-1)^2+(y+2)^2 =20$.

Show that the point (-3,0) lies on the circle:

$$(-3-1)^2+(0+2)^2 =20$$ $$(-4)^2+2^2 =20$$ $$16+4=20$$ $$20=20$$

The coordinates (-3,0) satisfy the equation. The point lies on the circle.

(b) (i) The equation of a line parallel to the given line $2x+2y-7=0$ is $2x+2y+k=0$, because parallel lines have the same slope and different intercepts.

Since the line CD has x-intercept equal to -3, the point (-3;0) satisfy the equation of CD:

$$2\cdot(-3)+2\cdot0+k=0$$ $$-6+k=0$$ $$k=6$$

The equation of the line CD is $2x+2y+6=0$.

(ii) Find the coordinates of the points of intersection of the line CD with the circle:

$$\begin{cases} 2x+2y+6=0 \\ (x-1)^2 +(y+2)^2 =20 \end{cases}$$ $$\begin{cases} y=-x-3 \\ (x-1)^2 +(-x-3+2)^2 =20 \end{cases}$$ $$\begin{cases} y=-x-3 \\ (x-1)^2 +(-x-1)^2 =20 \end{cases}$$ $$x^2 -2x+1+x^2 +2x+1=20$$ $$2x^2 =18$$ $$x^2=9$$ $$x=3,x=-3$$

If $x=3$ , then $y=-3-3=-6$.

If $x=-3$ , then $y=-(-3)-3=3-3=0$

The coordinates of the points of intersection of the line CD with the circle are (3,-6), (-3;0)

(iii) Find  the equation of the tangent of the circle at the point (-3,0). Determine the slope of the radius: $m_r=\frac{y_1-y_2}{x_1-x_2}$, where $(x_1,y_1)$ is the center of the circle, $(x_2,y_2)$ is the point on a circle.

$$m_r =\frac{-2-0}{1+3}=\frac{-2}{4}=-\frac{1}{2}$$

The slope of the tangent line must be perpendicular to the slope of the radius, so the slope of the line is $m_T=-\frac{1}{m_r}=2$

Use the point/slope form $y-y_1=m_T(x-x_1)$ at the point (-3,0):

$$y=2(x+3)$$ $$y=2x+6$$