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If $x^2 + y^2 - x - 2y - 5 = 0$ is a circle equation and $(-1,-1),(2,3)$ are two opposite end of a diameter of this circle. Find the two opposite end coordinates of the perpendicular diameter.

Hosam Hajeer
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Mohamed Asem
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  • First write down the circle's equation, locate its center and radius...and then we shall talk. – DonAntonio Mar 03 '23 at 08:27
  • With the given endpoints you can find the midpoint, the slope and the diameter. With that info, can you compute the perpendicular slope, and find two points that are half the diameter away from the center along the perpendicular line? Either algebraically or graphically. – Rócherz Mar 03 '23 at 08:44
  • Yes, I can do that. – Mohamed Asem Mar 03 '23 at 08:54
  • The midpoint of the diameter equals the two coordinates of the circle equation equals (1/2, 1) the length of the given diameter is 5/2 length units and the slope of these two points is 4/3 and that means that the perpendicular slpoe is 3/4 what after should I do? – Mohamed Asem Mar 03 '23 at 09:00
  • The product of the slopes of two perpendicular lines is $ \ \mathbf{-1} \ . \ $ So the perpendicular slope is...? You know the center of the circle, so you can get an equation for the line that the perpendicular diameter lies on. Insert this equation into the circle equation to find where this line intersects the circle. –  Mar 03 '23 at 09:15
  • Sorry I made a small mistake I mean -3/4 and not 3/4. – Mohamed Asem Mar 03 '23 at 09:17
  • When you rotate a vector $(u, v)$ by $90^\circ$ counterclockwise you get the vector $(-v, u)$ and when you rotate $90^\circ$ clockwise you get the vector $(v, -u) $. Then center of the circle is $(\frac{1}{2}, 1)$, so the vector extending from the center to $(2, 3)$ is $(\frac{3}{2}, 2)$. Rotate this last vector clockwise and counterclockwise to get $(2, - \frac{3}{2})$ and $(- 2, \frac{3}{2}) $ , then add the center coordinates to get the end points of the perpendicular diameter: $( \frac{5}{2}, -\frac{1}{2}) $ and $(- \frac{3}{2} , \frac{5}{2} ) $ – Hosam Hajeer Mar 03 '23 at 09:21
  • Yes, you are right sir, thank you very much. – Mohamed Asem Mar 03 '23 at 09:27

1 Answers1

2

Method 1:

The equation of the circle can be written as $$(x-\frac{1}{2})^2 + (y-1)^2 = (\frac{5}{2})^2$$ Hence, the center of the circle is $(\frac{1}{2}, 1)$

A line passing through $(-1, -1)$ and $(2, 3)$ has a slope of $\frac{4}{3}$

The slope of the perpendicular diameter will be the negative reciprocal of $\frac{4}{3}$, that is $-\frac{3}{4}$

A line passing through the center of the circle $(\frac{1}{2}, 1)$, with a slope of $-\frac{3}{4}$ has an equation $$y = -\frac{3}{4}x + \frac{11}{8}$$ Substitute this into the equation of the circle to get $$x^2 - x - \frac{15}{4} = 0$$ $$\implies x \in \left\{\frac{5}{2}, -\frac{3}{2}\right\}$$ $$\implies y \in \left\{-\frac{1}{2}, \frac{5}{2}\right\}$$ The required coordinates are $(\frac{5}{2},-\frac{1}{2})$ and $(-\frac{3}{2},\frac{5}{2})$

Method 2:

The center of the circle is the midpoint of $(-1, -1)$ and $(2, 3)$, that is $(\frac{1}{2}, 1)$

The length of the circle's diameter is the distance between $(-1, -1)$ and $(2, 3)$, that is $5$

The slope of the perpendicular diameter will be the negative reciprocal of $\frac{4}{3}$, that is $-\frac{3}{4}$

The two endpoints of the diameter must be of the form $(\frac{1}{2}+t, 1-\frac{3}{4}t)$ and $(\frac{1}{2}-t, 1+\frac{3}{4}t)$

The distance between the endpoints should be equal to the diameter of the circle $$\sqrt{(2t)^2 + (\frac{3}{2}t)^2} = 5$$ $$\implies t = 2$$

The required coordinates are $(\frac{5}{2},-\frac{1}{2})$ and $(-\frac{3}{2},\frac{5}{2})$

Amogh
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