C is a circle with centre the origin - Edexcel - GCSE Maths - Question 1 - 2020 - Paper 3

Using the point-slope form of a line, the equation of the tangent can be expressed as:

$y - y_1 = m(x - x_1)$

Choosing point (0, 10):

$y - 10 = \frac{1}{2}(x - 0)$

Thus,

$y = \frac{1}{2}x + 10.$

Since the circle is centered at the origin (0, 0) and the radius (r) is the distance from the center to the point where the tangent touches the circle, we can use the formula for the distance from a point to a line:

For the line given by: $Ax + By + C = 0$ where $A = \frac{1}{2}, B = -1, C = -10$,

The distance (d) from the origin to the line is given by:

$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

Substituting in the values:

$d = \frac{|\frac{1}{2} \cdot 0 - 1 \cdot 0 - 10|}{\sqrt{(\frac{1}{2})^2 + (-1)^2}} = \frac{10}{\sqrt{\frac{1}{4} + 1}} = \frac{10}{\sqrt{\frac{5}{4}}} = \frac{10 \cdot 2}{\sqrt{5}} = \frac{20}{\sqrt{5}} = 4\sqrt{5}.$

The equation of a circle with radius r and center at the origin is given by:

$x^2 + y^2 = r^2$.

Substituting the radius we found:

$r^2 = (4\sqrt{5})^2 = 16 \cdot 5 = 80.$

Thus, the equation of the circle C is:

$x^2 + y^2 = 80.$