Question

A circle passes through the points (-1, 1), (0, 6), and (5, 5). Find the points on this circle the tangents at which are parallel to the straight line joining the origin to its centre.

Answer 1

Hint: We need to find the equation of the circle using ${x^2} + {y^2} + 2gx + 2fy + c = 0$. Determine the values of g, f and c. From this equation we can the centre and hence the equation of the circle. Equation of line passing through the centre and perpendicular to line joining center to origin is
$y - {y_1} = - \dfrac{1}{m}(x - {x_1})$. Tangents will be parallel to the line joining the origin to the centre at the points somewhere at circle.

Complete step by step answer:
Let the equation of circle be${x^2} + {y^2} + 2gx + 2fy + c = 0$
It passes through (-1, 1)
$ \Rightarrow $1 + 1 -2g +2f + c =0
$ \Rightarrow $-2g + 2f +c = -2 … (1)
It passes through (0, 6)
$ \Rightarrow $0 + 36 + 0 + 12f + c = 0
$ \Rightarrow $12f + c = -36 … (2)
It passes through (5, 5)
$ \Rightarrow $25 + 25 + 10g + 10f +c =0
$ \Rightarrow $10g + 10f + c = -50 … (3)
Subtracting (2) from (1) we get
-2g - 10f = 34 … (4)
Subtracting (3) from (2) we get
-10g + 2f = 14 … (5)

Solving (4) and (5)
$ \Rightarrow $g = -2, f = -3
Substituting g and f in (1) we get c = 0
So the equation of circle is
 ${x^2} + {y^2} + 4x - 6y = 0$… (6)
Centre of the circle is $({x_1},{y_1})$= (2, 3)
Slope of line joining centre to origin, m =$\dfrac{{3 - 0}}{{2 - 0}}$=$\dfrac{3}{2}$
Equation of line passing through the centre and perpendicular to line joining center to origin is
$y - {y_1} = - \dfrac{1}{m}(x - {x_1})$
$y - 3 = - \dfrac{2}{3}\left( {x - 2} \right)$
3y – 9 = -2x + 4
2x + 3y =13 … (7)
Tangents will be parallel to the line joining the origin to the centre at the points where (7) cuts the circle
Substituting y from (7) in (6) we get
$ \Rightarrow {x^2} + {\left( {\dfrac{{13 - 2x}}{3}} \right)^2} + 4x - 6\left( {\dfrac{{13 - 2x}}{3}} \right) = 0$
$ \Rightarrow {x^2} + \left( {\dfrac{{169 + 4{x^2} - 52x}}{9}} \right) + 4x - 2\left( {13 - 2x} \right) = 0$
$ \Rightarrow \dfrac{{9{x^2} + 169 + 4{x^2} - 52x + 36x - 18\left( {13 - 2x} \right)}}{9} = 0$
$ \Rightarrow 9{x^2} + 169 + 4{x^2} - 52x + 36x - 234 - 36x = 0$
$ \Rightarrow 13{x^2} - 52x - 65 = 0$
$ \Rightarrow {x^2} - 4x - 5 = 0$
$ \Rightarrow {x^2} - 5x + x - 5 = 0$
$ \Rightarrow x(x - 5) + 1(x - 5) = 0$
$ \Rightarrow (x + 1)(x - 5) = 0$
Therefore x = -1, 5
Substituting x in (7) we get y = 5, 1
So the point of contact of tangents are (-1, 5) and (5, 1)

Note: A tangent is a straight line that touches the circumference of a circle at only one place. The product of the gradient of the radius and the gradient of the tangent line is equal to -1. The equation of the circle is${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$, where r is the radius.