MATHS

Asked by KaitlynFirst we need the equation of our line: We have the y-intercept (which is (0,10)). We then need to find the gradient of our line which is: Change in y / change in x = 10/20 = 0.5 so the equation of our line is: y = 0.5x + 10 We know the centre of our circle is the origin: so it is of the form: x^2 + y^2 = r^2 (where r is the radius) As we need the line to meet our circle at one point, we can substitute the equation for our straight line in for y. the equation of our circle is now: x^2 + (0.5x^2 + 10)^2 = r^2 x^2 + (0.5x + 10)(0.5x + 10) = r^2 we can then expand and simplify the brackers to leave us with: 1.25x^2 + 10x + 100 = r^2 1.25x^2 + 10x + 100 - r^2 = 0 As this is a tangent (and so it only meets the circle at one point) This equation should only have one solution. We can then set our discriminant to zero (as this means our equation will have one solutuon for r) and from there we can find a value for r that satisfies this condition. b^2 - 4ac = 0 10^2 - 4(1.25)(100 - r^2) = 0 If we expand our brackets and rearrange this, we will find that r^2 = 80 Therefore the equation of our circle is: x^2 + y^2 = 80 If needed, you can book a lesson for further understanding