Question

What is the equation of the line tangent to `(x-6)^2+(y-9)^2=34` at the point `(9,4)` ?

Answer 1

`y=3/5x-7/5`

Explanation:

`"the equation of a circle in standard form is"`

`•color(white)(x)(x-a)^2+(y-b)^2=r^2`

`"where "(a,b)" are the coordinates of the centre and r "`
`"is the radius"`

`(x-6)^2+(y-9)^2=34" is in standard form"`

`rArr"centre "=(6,9)`

`• " the tangent to the circle and radius are at right angles"`
`"at the point of contact"`

`m_(color(red)"radius")=(y_2-y_1)/(x_2-x_2)`

`"let "(x_1,y_1)=(6,9)" and "(x_2,y_2)=(9,4)`

`rArrm_(color(red)"radius")=(4-9)/(9-6)=(-5)/3=-5/3`

`•color(white)(x)m_(color(red)"radius")xxm_(color(red)"tangent")=-1`

`rArrm_(color(red)"tangent")=-1/(-5/3)=3/5`

`•color(white)(x)y-y_1=m(x-x_1)larrcolor(blue)"point-slope form"`

`"with "m=3/5" and "(x_1,y_1)=(9,4)`

`y-4=3/5(x-9)`

`y-4=3/5x-27/5`

`rArry=3/5x-7/5larrcolor(blue)"in slope-intercept form"`

Answer 2

`y = 1/5(3x-7)`

Explanation:

`(x-6)^2+(y-9)^2=34`

This is the equation of a circle centred at the point `(6,9)` with a radius of `sqrt34`

The slope at any point `(x,y)` on a circle centred at the origin is `-x/y`

Hence, the slope of any point on a circle centred at `(6,9)` is `-(x-6)/(y-9)`

To find the slope `(m)` of the tangent to the given circle at the point `(9,4)`

`m= -(9-6)/(4-9) = -3/-5 = 3/5`

The equation of a straight line with slope `m` passing through the point `(x_1,y_1)` is:

`(y-y_1) = m(x-x_1)`

Here our tangent line has slope `3/5` and passes through the point `(9,4)`

Hence, `(y-4) = 3/5(x-9)`

`y = 3/5x -27/5+20/5`

`y= 3/5x -7/5`

`y= 1/5(3x-7)`

We can see this result from the graphic below showing the given circle and our tangent at `(9,4)`

graph{((x-6)^2+(y-9)^2-34)(-y+1/5(3x-7))=0 [-32.47, 32.5, -16.25, 16.22]}