What is the tangent line equation that touches the circle of x squared plus y squared -6x plus 4y -7 equals 0 at the point of 1 2 on the Cartesian plane showing work?

Answer 1

The line tangent to a circle is perpendicular to the radius from the centre of the circle to the point of contact.

A circle with centre (X, Y) and radius r has equation:

(x - X)² + (y - Y)² = r²

Re-arranging x² + y² -6x + 4y -7 = 0 into this form allows us to find the centre of the circle. This requires completing the square in both x and y:

x² + y² - 6x + 4y - 7 = 0

→ x² -6x + 9 - 9 + y² + 4y + 4 - 4 - 7 = 0

→ (x - 3)² - 9 + (y + 2)² - 4 - 7 = 0

→ (x - 3)² + (y + 2)² - 20 = 0

→ (x - 3)² + (y + 2)² = 20

→ the centre of the circle is at (3, -2).

The slope of the radius between this centre and the point of touch at (1, 2) is given by:

m = change_in_y / change_in_x = (2 - -2)/(1 - 3) = 4/-2 = -2

The slope m' of a line perpendicular to a line of slope m is such that mm' = -1

→ m' = -1/m

→ the slope of the tangent m' is m' = -1/-2 = 1/2

The equation of a line through point (x0, y0) with slope m is given by:

y - y0 = m(x - x0)

Thus the equation of the tangent (with slope 1/2) to the circle at (1, 2) is:

y - 2 = ½(x - 1)

→ 2y - 4 = x - 1

→ 2y = x + 3

→ 2y - x = 3

(or y = x/2 + 1.5)