Question

if the centre of a circle (2,3) and a tangent is x+y=1 then the equation of the circle is

Answer 1

The tangent is x + y = 1 which is the same as y = -x + 1.

So, gradient of tangent = -1.

We know that radius meets tangent at a right angle.

So, the radius that meets this tangent at the point of contact has gradient -1/-1 = 1.

So, the equation of this radius is :

$y - 3 = 1(x - 2) \implies y = x + 1.$

Find the point of contact of the radius and tangent:

$x + y = 1 \implies x + x + 1 = 1 \implies x = 0.$

$x = 0 \implies y = 1.$

So, (0, 1) lies on the circumference of the circle.

Hence the radius of the circle is:

$r = \sqrt{(2-0)^{2} + (3 - 1)^{2}} = \sqrt{8}.$

Hence, the equation of the circle is:

$\underline{\underline{(x - 2)^{2} + (y - 3)^{2} = 8}}.$

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