Mursalin's circle

Use Power of a Point. Extend OT to become a diameter of the circle, with points X and Y on the circle such that X is closer to T and Y is closer to O. Letting "r" be the radius, we have TX = r-17 and TY = r+17, so by PoP, (AT)(TB)=(XT)(TY), which translates to (7)(17) = (r-17)(r+17). The value of N is r^2, which we solve to be 408 .

Let $C$ be a point on $AB$ so that $OC$ perpendicular to $AB$ .

$AC=\frac12 AB=12$

$CT=AC-AT=12-7=5$

$OC^2=OT^2-CT^2=17^2-5^2=264$

Let $r$ be radii of $\Gamma$

$r^2=OB^2=OC^2+BC^2=264+144=408$

Area of $\Gamma=\pi r^2=408\pi$

Thus, $N=408$

Let the perpendicular from $O$ on $AB$ meet $AB$ at $M.$

We know that $M$ is the mid point of $AB.$

$\Rightarrow AM = MB = 12$ and $MT =5$

In right triangle $OMT$

$OM^2 = OT^2 - MT^2$

$\Rightarrow OM^2 = 17^2 - 5^2 = 264$

In right triangle $OMB$

$OB^2 = OM^2 + BM^2 = 264 + 12^2 = 408$

Area of the circle = $\pi * OB^2 = 408\pi$

$\Rightarrow N = 408$

AB - AT = 24 - 7 = 17 = TB

This shows that BTO is a isosceles triangle (because OT and BT are equal).

Since OA and OB are both radiuses of the triangle, OAB is also isosceles.

Since angle OBA and OBT are the same angle to each other, they are congruent to each other. By the properties of isosceles triangles, OBA is also congruent to OAB and OBT is congruent to TOB. Therefore OAB is congruent to TOB.

Because there are two angles are congruent between triangles BTO and OAB, the third is as well, and the triangles are similar. so the two equal sides of OAB (the radii) are proportional to the 17, and the radius that makes up the third side of BTO is proportional to the third side of OAB (AB, which is 24).

$\frac{r}{24} = \frac{17}{r}$

$r^{2} = 17 \times 24$

since the area of a triangle is $πr^{2}$ , N = $r^{2}$ , so N = 408.

Extend the segment $OT$ to create a diameter that meets the circle at points $C$ and $D$ with $T$ being between $C$ and $O$ . Calling $CT=x$ , you can use the intersecting chords theorem to solve for $x$ . $x(34+x)=17*7$ $x^2+34x-119=0$ $x=\frac{-34 \pm \sqrt{1632}}{2}$ $x=-17 + 2\sqrt{102}$ The radius of the circle is $x+17$ or $2\sqrt{102}$ and the area will be $(2\sqrt{102})^2\pi$ or $408\pi$

We have two segments, AB and the diameter (we can call FG). The meet each other in T point. So, we know AT=7 and TB = 24 - 7 = 17. We also know FT= r+17 and TG= r-17. Using the fact that the product of the two parts of the chord is equal to the product of the two parts of the other chord, we have: AT . TB = FT . FG => 7 . 17 = (r+17)(r-17) => 119 = r² - 17² ...... *[because (a+b)(a-b) = a²-b²] => 119 = r² -289 => 408 = r² . Once the area of a circle is given by π . r² , the area of Γ can be written as 408π, so 408 =N.

Firstly, $N =r^2$ so we just need to work out the radius.

Note that triangle $ABO$ is an isosceles, which means that the altitude passing through $O$ is also the perpendicular bisector of $AB$ .

Let the foot of this perpendicular be $M$ . Clearly $M$ lies on the same side of $T$ as $B$ , so we now know that $TM = 17-12 = 5$

$MTO$ is a right angle triangle, with sides $5, 17,$ and $h$ , where $h$ is the height of triangle $ABO$ . $h^2 = 17^2-5^2$

Now going back to $ABO$ , we can see that $AOM$ is also a right angle triangle with sides $12,h,$ and $r$ , where $r$ is the radius of the circle). Now we have $r^2 = 17^2-5^2+12^2 = 408$

So $N=408$

Drop the altitude from $O$ to $AB$ and call that $C$ . We know the line that is the perpendicular bisector of a chord passes through the center. This implies at $AC=BC=12 \implies TC=5$ . Use the Pythagorean Theorem to find $OT$ , which is $\sqrt{264}$ . Use it again to find the radius: $264+144=408=r^2$ .

That implies our answer is $\boxed{408}$