What is the tangent equation of the circle 2x2 plus 2y2 -8x -5y -1 equals 0 when it touches the point 1 -1 on the Cartesian plane?

Answer 1

There are four steps required in solving this:

1. Find the centre of the circle by rearranging the equation into the form (x - Xo)² + (y - Yo)² = r² where (Xo, Yo) is the centre of the circle and r is the radius of the circle - this is done by completing the square for each of x and y;
2. Find the slope m of the radius between the point of contact (Xc, Yc) and the centre (Xo, Yo) of the circle which is given by m = change_in_y/change_in_x = (Yc - Yo)= (Xc - Xo) = (Yo - Yc)/(Xo - Xc);
3. The tangent is perpendicular to the radius, so its slope m' can be found from the relation mm' = -1 → m' = -1/m;
4. Find the equation of the tangent passing through the point (Xc, Yc) with slop m' by using the slope-point equation of a line: Y - Yc = m'(x - Xc)

Have a go at finding the solution before reading on.

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Circle equation: 2x² + 2y² - 8x - 5y - 1 = 0

Divide all terms by 2: x² + y² - 4y - 2.5 - 0.5 = 0

Completing the squares: (x - 2)² + (y - 1.25)² = 6.0625

Center of circle: (2, 1.25)

Point of contact: (1, -1)

Slope of radius: 9/4

Slope of tangent line: -4/9

Tangent equation: y - -1 = -(4/9)(x - 1)

→ 9y = -4x - 5

→ 9y + 4x + 5 = 0