Question

The equation of a circle is (x - 3)2 + (y + 2)2 = 25. The point (8, -2) is on the circle.

What is the equation of the line that is tangent to the circle at (8, -2)?

Answer 1

Hello,

O=(3,-2) is the center of the circle and 5 its radius.

P=(8,-2)
The line OP is parallele to ox axis so then tangent is parallele to oy axis
Its equation is x=8.

Answer 2

The equation of the line that is tangent to the circle at (8, -2) is:

`x=8`

Explanation:

We know that the line which is tangent at a point on a circle is perpendicular to the line joining the center of the circle and that point.

Here we are given the equation of a circle as:

`(x-3)^2+(y+2)^2=25`

The center of the circle is at: (3,-2)

( Since, the standard form of a circle with center at (h,k) and radius r is given by:

`(x-h)^2+(y-k)^2=r^2` )

Also, the equation of a line joining (3.-2) and (8,-2) is given by:

`y-(-2)=(-2-(-2))/(8-3)* (x-3)\\\\i.e.\\\\y+2=(-2+2)/(5)* (x-3)\\\\i.e.\\\\y+2=(0)/(5)* (x-3)\\\\i.e.\ y+2=0\\\\i.e.\ y=-2`

Also, we know that the slope of this line is zero.

Also, we know that if two lines are perpendicular with slope m and m' respectively then,

`m\cdot m'=-1\\\\i.e.\\\\m'=(-1)/(m)`

Hence, we get that the slope of the tangent line is:

`(-1)/(0)`

Also, we know that:

The equation of a line with given slope m' and a passing through point (a,b) is given by:

`y-b=m'(x-a)`

Here (a,b)=(8,-2)

and
`m'=(-1)/(0)`

i.e. the equation of the tangent line is:

`y-(-2)=(-1)/(0)(x-8)\\\\i.e.\\\\y+2* 0=-1(x-8)\\\\i.e.\\\\-1* (x-8)=0\\\\i.e.\\\\(x-8)=0\\\\i.e.\\\\x=8`