Question

Show that `CM` and `RQ` are perpendicular ?

Answer 1

See below.

Explanation:

Given `R=(3a,a)` and `Q=(a,-3a)` the mid point of `RQ` is given by

`M=1/2(R+Q)=(2a,-a)`

The circle's center is `C=(0,0)`

so `CM=(2a,-a)-(0,0)=(2a,-a)` and

`RQ = R-Q =(2a,4a)`

Finally

`RQ cdot CM = (2a,4a) cdot (2a,-a) = 4a^2-4a^2=0`

Attached a figure

Answer 2

The x coordinate of the chord goes from `3a` to `a`, therefore, the x coordinate of the midpoint is halfway between them, 2a.

The y coordinate of the chord goes from `a` to `-3a`, therefore, the y coordinate of the midpoint is halfway between them, -a.

The coordinates of the midpoint are `M(2a,-a)`

The slope, `m_1`, of the chord RQ is:

`m_1 = (-3a - a)/(a - 3a)`

`m_1 = (-4a)/( -2a)`

`m_1 = 2`

The equation of the circle `x^2 + y^2 = 10a^2` is a special case of the more general equation `(x - h)^2+(y-k)^2 = r^2` where the center is the origin `(0,0)`.

The slope, `m_2`, of the line from the center to the midpoint is:

`m_2 = (-a - 0)/(2a - 0)`

`m_2 = (-a)/(2a)`

`m_2 = -1/2`

Please observe that:

`m_1m_2 = -1`

This shows that the two lines are perpendicular.