A circle has equation $x^2+y^2-4x+2y-20=0$. The point $P$ has coordinates $(t,t-3)$, where&nbs

Question

A circle has equation $x^2+y^2-4x+2y-20=0$. The point $P$ has coordinates $(t,t-3)$, where $t$ is a real number. Find the centre and radius of the circle.[2] Find the exact range of values of $t$ for which $P$ lies strictly inside the circle.[3]

Accepted Answer

(a) Centre $(2,-1)$, radius $5$. (b) $2-\dfrac{5\sqrt2}{2}<t<2+\dfrac{5\sqrt2}{2}$.

A moving point inside a circle

A moving point inside a circle

Solution:

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