find the segment length indicated. assume that lines which appear to be tangent are tangent.
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problem 3
problem 4

Question

Accepted Answer

### Problem 3
# Explanation:
## Step 1: Recall the tangent - secant theorem
The tangent - secant theorem states that if a tangent segment and a secant segment are drawn from an external point to a circle, then the square of the length of the tangent segment is equal to the product of the lengths of the entire secant segment and its external part. Let the length of the tangent be $t = 7.2$, the length of the external part of the secant be $x$ and the length of the entire secant be $x + d$, where $d=4.8$ (the diameter of the circle). But we can also use the Pythagorean theorem here because the radius is perpendicular to the tangent at the point of contact. Let the radius be $r=\frac{4.8}{2} = 2.4$, the length of the tangent is $7.2$, and let the length of the secant segment from the external point to the point of intersection with the circle (the part we need to find) be $y$. Wait, actually, the line from the external point to the center forms a right triangle with the tangent and the radius. Wait, no, the tangent is perpendicular to the radius, so if we consider the right triangle formed by the radius ($r = 2.4$), the tangent ($t=7.2$) and the line from the external point to the center (let's call it $L$). But actually, the secant here: the secant has a length from the external point to the far intersection point as $y + 4.8$ (where $y$ is the length from the external point to the near intersection point) and the external part is $y$, and the tangent is $7.2$. By the tangent - secant rule: $t^{2}=y(y + 4.8)$. Wait, no, maybe I made a mistake. Let's re - consider. The diameter is $4.8$, so the radius $r = 2.4$. The tangent is $7.2$, and the secant: the length from the external point to the near point is $y$ (the one we need to find) and from the near point to the far point is $4.8$. So by the tangent - secant theorem: $7.2^{2}=y(y + 4.8)$. Wait, no, the formula is $t^{2}=a(a + b)$, where $t$ is the tangent, $a$ is the external segment, and $a + b$ is the entire secant. But in this case, the secant passes through the center, so the length of the secant from the external point to the far point is $y+4.8$ and the external segment is $y$. So $7.2^{2}=y(y + 4.8)$. Wait, but maybe it's easier to use the Pythagorean theorem. Let the distance from the external point to the center be $d$. Then $d^{2}=r^{2}+t^{2}$, and also $d^{2}=(y + r)^{2}-r^{2}$? No, wait, the radius is $r = 2.4$, the tangent is $t = 7.2$, and the length of the secant from the external point to the point of intersection (the part we need to find) is $y$. Wait, actually, the line from the external point to the center: let's call the external point $P$, the point of tangency $T$, the center $O$, and the points of intersection of the secant with the circle $A$ (near) and $B$ (far), so $AB = 4.8$ (diameter), $PT = 7.2$ (tangent), $PA=y$, $PB=y + 4.8$. By the tangent - secant theorem: $PT^{2}=PA\times PB$, so $7.2^{2}=y(y + 4.8)$. Let's solve for $y$:

$51.84=y^{2}+4.8y$

$y^{2}+4.8y - 51.84 = 0$

We can use the quadratic formula $y=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, where $a = 1$, $b = 4.8$, $c=- 51.84$

$\Delta=b^{2}-4ac=(4.8)^{2}-4\times1\times(-51.84)=23.04 + 207.36=230.4$

$\sqrt{\Delta}=\sqrt{230.4}\approx15.18$

$y=\frac{-4.8\pm15.18}{2}$

We take the positive root: $y=\frac{-4.8 + 15.18}{2}=\frac{10.38}{2}=5.19$? Wait, that can't be right. Wait, maybe I messed up the theorem. The correct formula for tangent and secant: if a tangent of length $t$ and a secant of length $s$ (where $s$ is the length from the external point to the far intersection point) and the external part of the secant is $e$ (from external point to near intersection point), then $t^{2}=e\times s$. In this case, the secant passes through the center, so the length of the secant from the external point to the far point is $e + 4.8$ (since the diameter is $4.8$), and the external part is $e$ (the length we need to find). The tangent is $7.2$. So $7.2^{2}=e(e + 4.8)$

$51.84=e^{2}+4.8e$

$e^{2}+4.8e - 51.84 = 0$

Multiply through by 100 to eliminate decimals: $100e^{2}+480e-5184 = 0$

Divide by 8: $12.5e^{2}+60e - 648=0$? No, better to use the quadratic formula as before. Wait, maybe I made a mistake in the diagram. Wait, the line that is a secant: is it a secant or is it a line that intersects the circle at two points, one of which is the end of the diameter? Wait, the diameter is $4.8$, so the radius is $2.4$. The tangent is $7.2$, and the other line is a secant that goes from the external point, through the circle, with the diameter being a part of the secant. So the length of the secant from the external point to the near point is $x$ (the unknown), and from the near point to the far point is $4.8$. So by the tangent - secant theorem: $7.2^{2}=x(x + 4.8)$

$x^{2}+4.8x-51.84 = 0$

Using the quadratic formula $x=\frac{-4.8\pm\sqrt{4.8^{2}+4\times51.84}}{2}=\frac{-4.8\pm\sqrt{23.04 + 207.36}}{2}=\frac{-4.8\pm\sqrt{230.4}}{2}=\frac{-4.8\pm15.18}{2}$

We take the positive root: $x=\frac{-4.8 + 15.18}{2}=\frac{10.38}{2}=5.19$? No, that seems odd. Wait, maybe the diagram is such that the line from the external point to the point of intersection with the circle (the secant) and the tangent form a right triangle with the line from the external point to the center. Wait, the radius is $2.4$, the tangent is $7.2$, and the distance from the external point to the center is $d$. Then $d^{2}=2.4^{2}+7.2^{2}=5.76 + 51.84 = 57.6$. Then the length of the secant from the external point to the far point (through the center) is $d + 2.4$? No, wait, the center is on the secant. So the distance from the external point to the center is $d$, and the length of the secant from the external point to the near point is $d - 2.4$ and from the near point to the far point is $4.8$ (so from the external point to the far point is $d - 2.4+4.8=d + 2.4$). But by the tangent - secant theorem, $7.2^{2}=(d - 2.4)(d + 2.4)=d^{2}-2.4^{2}$. But we know $d^{2}=57.6$, so $7.2^{2}=57.6 - 5.76=51.84$, which is true. But we need to find the length of the secant from the external point to the near point, which is $d - 2.4$. Since $d^{2}=57.6$, $d=\sqrt{57.6}\approx7.59$, then $d - 2.4\approx7.59 - 2.4 = 5.19$. Wait, but maybe there is a miscalculation. Wait, maybe the formula is different. Wait, the tangent - secant formula is $t^{2}=a\times b$, where $a$ is the external segment and $b$ is the entire secant. If the secant passes through the center, then $b=a + d$ (where $d$ is the diameter). So $t^{2}=a(a + d)$. So $7.2^{2}=a(a + 4.8)$. Let's solve for $a$:

$a^{2}+4.8a-51.84 = 0$

Using the quadratic formula $a=\frac{-4.8\pm\sqrt{4.8^{2}+4\times51.84}}{2}=\frac{-4.8\pm\sqrt{23.04 + 207.36}}{2}=\frac{-4.8\pm\sqrt{230.4}}{2}=\frac{-4.8\pm15.18}{2}$

$a=\frac{-4.8 + 15.18}{2}=5.19$ or $a=\frac{-4.8 - 15.18}{2}=-9.99$ (discard the negative). So the length of the segment we need is approximately $5.2$? Wait, maybe I made a mistake in the diagram interpretation. Alternatively, maybe the line is a secant and the tangent, and the triangle is a right triangle? Wait, no, the tangent is perpendicular to the radius, so the triangle formed by the radius, the tangent, and the line from the external point to the center is a right triangle. The line from the external point to the center has length $L$, the radius $r = 2.4$, the tangent $t = 7.2$, so $L=\sqrt{r^{2}+t^{2}}=\sqrt{2.4^{2}+7.2^{2}}=\sqrt{5.76 + 51.84}=\sqrt{57.6}\approx7.59$. Then the length of the secant from the external point to the near point is $L - r\approx7.59 - 2.4 = 5.19\approx5.2$. But maybe the problem expects an exact value. Wait, $7.2=\frac{36}{5}$, $4.8=\frac{24}{5}$. Then $t^{2}=(\frac{36}{5})^{2}=\frac{1296}{25}$, and $a(a + \frac{24}{5})=\frac{1296}{25}$. Multiply through by 25: $25a^{2}+120a - 1296 = 0$. Divide by 4: $6.25a^{2}+30a - 324 = 0$. Using quadratic formula: $a=\frac{-30\pm\sqrt{900+4\times6.25\times324}}{2\times6.25}=\frac{-30\pm\sqrt{900 + 8100}}{12.5}=\frac{-30\pm\sqrt{9000}}{12.5}=\frac{-30\pm94.868}{12.5}$. Positive root: $\frac{64.868}{12.5}\approx5.19$. So approximately $5.2$. But maybe the diagram is different. Wait, maybe the secant is not passing through the center? No, the dot is the center, so the line is a diameter.

### Problem 4
# Explanation:
## Step 1: Recall the tangent - radius theorem and Pythagorean theorem
The tangent to a circle is perpendicular to the radius at the point of contact. So the triangle formed by the tangent, the radius, and the line from the external point to the center is a right triangle. The diameter of the circle is $2\times7.5 = 15$ (since the radius is $7.5$). Let the length of the tangent be $x$ (the unknown), the length of the line from the external point to the end of the chord (the non - tangent side) be $17$, and the length of the diameter be $15$. Wait, the chord length is $17$? No, the line segment of length $17$ is from the external point to the point on the circle (the end of the chord). Wait, the radius is $7.5$, so the diameter is $15$. The triangle is a right triangle with hypotenuse equal to the distance from the external point to the center, one leg is the radius ($7.5$) and the other leg is the tangent ($x$), and the other leg (the chord - related side) is $17$? Wait, no. Wait, the line from the external point to the point on the circle (the end of the chord) is $17$, the radius is $7.5$, and the tangent is $x$. Wait, the distance from the external point to the center: let's call the external point $P$, the point of tangency $T$, the center $O$, and the point on the circle (end of the chord) $A$. Then $OA = 7.5$ (radius), $PA = 17$, $PT=x$ (tangent), and $OT\perp PT$, so triangle $PTO$ is right - angled at $T$, and triangle $PAO$ is a triangle with sides $PA = 17$, $OA = 7.5$, and $PO$ (the hypotenuse of triangle $PTO$). Wait, no, $PO$ is the hypotenuse of triangle $PTO$, so $PO^{2}=x^{2}+7.5^{2}$. Also, in triangle $PAO$, we can use the Pythagorean theorem? Wait, no, $PA$ is not necessarily perpendicular to $OA$. Wait, the length of the chord: the chord length is not given, but the line from $P$ to $A$ is $17$, and $OA = 7.5$, and $PT$ is tangent. Wait, maybe the line from $P$ to $A$ is a secant? No, it's a tangent? Wait, no, the diagram shows a tangent (the one with the question mark) and a secant? Wait, no, the line with length $17$ is a chord? No, the dot is the center, so the line from the center to $A$ is a radius ($7.5$), so the diameter is $15$. The line from the external point to $A$ is $17$, and the tangent is $x$. So by the Pythagorean theorem in the right triangle formed by the tangent ($x$), the radius ($7.5$) and the line from the external point to the center ($d$): $d^{2}=x^{2}+7.5^{2}$. Also, the line from the external point to $A$ is $17$, and $OA = 7.5$, so in triangle $PAO$, $d^{2}=17^{2}-7.5^{2}$ (wait, no, that would be if $PA$ is perpendicular to $OA$, but is it? Wait, the tangent is perpendicular to the radius, but $PA$ is a chord? No, the diagram: the tangent is the side with the question mark, the line of length $17$ is from the external point to a point on the circle (not the point of tangency), and the radius is $7.5$. Wait, maybe the triangle is a right triangle with the tangent ($x$), the line of length $17$, and the diameter? No, the diameter is $15$. Wait, the correct approach: the tangent is perpendicular to the radius, so we have a right triangle where one leg is the tangent ($x$), one leg is the radius ($7.5$), and the hypotenuse is the distance from the external point to the center ($d$). Also, the

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Explanation:

Problem 3

Step 1: Recall the tangent - secant theorem

Step 1: Recall the tangent - radius theorem and Pythagorean theorem

Answer:

Step 1: Recall the tangent - radius theorem and Pythagorean theorem