Solve 4x^2 = 70. round to the nearest hundredth. the solutions are ±

You have to use the discriminant b^2 -4ac =0 because lines that cut a circle only once are tangents.

Step-by-step explanation:

I think you can use the perpendicular gradient, but it'll be more tedious because then you'll have to prove that the line touching the circle is perpendicular to the line from the centre to the circle to the point of intersection. You'll have to form an equation using the coordinates of the centre of the circle and the point of intersection. You probably had a careless mistake/ calculation error in this process.

Instead, try

1. equating the line to the circle (which you did), then

2. let the coefficient of x^2 be a, the coefficient of x be b, and the constant terms be c.

3. simplify b^2-4ac

3. prove that b^2-4ac=0 because there is only one real and distinct root, hence there is only one point of intersection.