Point in, on or out of a circle

Question

Let $\;C\;$ be a circle with diameter $XY$ and let $P$ be a point not on the line $XY.$ Then

(1). If $P$ is on the circle $C,$ then $\angle XPY = 90^{\circ}$

(2). If $P$ is inside the circle $C,$ then $\angle XPY >90^{\circ}$

(3). If $P$ is outside the circle $C,$ then $\angle XPY < 90^{\circ}$

How to prove the statements (2) & (3)?

Any help would be appreciated. Thank you.

Answer 1

2) Produce $XP$ to meet the circle at $Q$. Then $\angle XQY=90^\circ$ and $\angle XPY>\angle XQY$.

3) Suppose that $XP$ meets the circle at $R$. Then $\angle XRY=90^\circ$ and $\angle XPY<\angle XRY$.

Answer 2

Use the following fact.

Let given $\Delta ABC$ and $D$ on the line $AC$ such that $C$ placed between $A$ and $D$. Prove that: $$\measuredangle BCD>\measuredangle A$$ and $$\measuredangle BCD>\measuredangle B.$$

Answer 3

(1) is just Thales' theorem. Then remember secant angles, which with one chord being the diameter means $\angle XPY = 90^\circ + \alpha/2$ for an interior point, and $\angle XPY = 90^\circ - \alpha/2$ for an exterior one.

Answer 4

In the triangle $XPY$ the sum of angles is $\pi$, so $p=\pi-(x+y)$

$(1)$

When $P$ is on the circle the angle is $\frac{\pi}2$

• because $[OP],[OX],[OY]$ are radius then $OPX$ and $OPY$ are isoceles in $O$ so $x=\widehat{XPO}$ and $y=\widehat{YPO}$ and $p=x+y\implies 2p=\pi$.

$(2)$

Now if you keep $P$ on the straight line issued for $O$ at constant angle with $(XY)$:

When $P_2$ is further away (out of the circle) than $P_1$ then $x_2>x_1$ and $y_2>y1$, thus $x_2+y_2>x_1+y_1\iff p_2<p_1$, this is simpler as that.

And reciprocally when $P_2$ is nearer (inside the circle) then $p_2>p_1$.