Question

A and B are the points of intersection of circle x^2+y^2-2x+4y-75=0 and its diameter which passes through the origin. Find the coordinate of A and B. Also find the equation of tangents at A and B.?

Answer 1

The points of intersection are `A = (5,-10)` and `B = (-3,6)`

The equations of the tangent lines are:

`y = 1/2(x-5)-10` and `y = 1/2(x--3)+6`

Explanation:

Given: `x^2+y^2-2x+4y-75=0" [1]"`

The standard Cartesian form for the equation of a circle is:

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

where `(h,k)` is the center and `r` is the radius

Add 75 to both sides of equation [1]:

`x^2+y^2-2x+4y=75`

Group the x terms and y terms together:

`x^2-2x+y^2+4y=75" [1.1]"`

Expand equation [2]:

`x^2-2hx+h^2+ y^2-2ky+k^2 = r^2" [2.1]"`

Add `h^2+k^2` to both sides of equation [1.1]:

`x^2-2x+h^2+y^2+4y+k^2=75+h^2+k^2" [1.2]"`

We set the second term in equation [2.1] equal to the second term in equation [1.2] to find the value of `h`:

`-2hx = -2x`

`h= 1`

We can set the fifth term in equation [2.1] equal to the fifth term in equation [1.2] to find the value of k:

`-2ky = 4y`

`k = -2`

The center of the circle is `(1,-2)`

The slope of the line that passes through the center and the origin is:

`m = (-2-0)/(1-0)`

`m =-2`

The equation of the line is:

`y = -2x" [3]"`

To find the x coordinates where the line intersects the circle, substitute equation [3] into equation [1]:

`x^2+(-2x)^2-2x+4(-2x)-75=0`

`5x^2-10x-75=0`

`x^2 - 2x - 15 = 0`

`(x-5)(x+3) = 0`

`x = 5` and `x = -3`

Use equation [3] to find the corresponding y coordinates:

`y = -10` and `y = 6`

The points of intersection are `A = (5,-10)` and `B = (-3,6)`

The tangent lines are perpendicular to the line describe by equation [3], therefore, the slope, `n`, of the tangent lines will be the negative reciprocal of the slope of equation [3]:

`n = -1/m`

`n = -1/-2`

`n = 1/2`

Use the point-slope form of the equation of a line to write the equations of the tangent lines:

`y = 1/2(x-5)-10` and `y = 1/2(x--3)+6`