MATHS

Asked by AmyFirst of all, we want to know the equation of our straight line: It passes through the points (-30 , 0) and (0, 10) therefore we can find the gradient of this line. Change in y/ change in x will give us 30/10 therefore our gradient is 3. This line also has a Y intercept of 10 as we can see. Therefore the equation of our straight line is y = 3x + 10 now for the equation of our circle... It is given that the circle passes through our origin therefore we can write the equation of our circle as: x^2 + y^2 = r^2 (Where the x^2 means x-squared.) Essentially what we wish we do now is find the radius of this circle so that it only touches our straight line once. We can do that firstly substituting in the value of y that we have in the first equation. y = 3x + 10 So we end up with: x^2 + (3x + 10)^2 = r^2 we can write this as x^2 + (3x + 10)^2 - r^2 = 0 we can expand this into: x^2 + 9x^2 + 60x + 100 - r^2 = 0 and then 10x^2 + 60x + 100 - r^2 = 0 As we only want our line to intersect the curve at one point, this equation should only have one solution for r. This is only true if the discriminant = 0 so we can say b^2 - 4ac = 0 and subsequently... 60^2 - 4(10)(100 - r^2) = 0 This is a quadratic equation, and we can solve this. 3600 - 4000 + 40r^2 = 0 -400 + 40r^2 = 0 400 = 40r^2 r^2 = 10 Therefore we can write the equation of our circle as x^2 + y^2 = 10 and this is our equation for C ☝🏾