Question

Consider the circle $x^2 + y^2 - 8x - 18y + 93 = 0$ with centre C and point P(2, 5) outside it. From the point P, a pair of tangents PQ and PR are drawn to the circle with S as mid point of QR. The line joining P to C intersect the given circle at A and B. Which of the following is/are true?

• A. The angle between two tangents from P is $tan^{-1}(\frac{4}{3})$
• B. CP is arithmetic mean of AP and BP
• C. Area of $\triangle PQR$ is $\frac{32}{5}$ unit²
• D. Equation of circumcircle of $\triangle PQR$ is $x^2 + y^2 - 6x - 14y + 53 = 0$.

Answer 1

Answer: (A), (B), (C), (D)

All the statements are true. (A), (B), (C), (D)

Answer 2

We are given the circle

$$x^2+y^2-8x-18y+93=0.$$

Step 1. Find the centre and radius.

Complete the square:

$$\begin{aligned} x^2 - 8x &= (x-4)^2-16,\\[1mm] y^2 - 18y &= (y-9)^2-81. \end{aligned}$$

Thus the equation becomes

$$(x-4)^2+(y-9)^2-16-81+93=0 \quad\Longrightarrow\quad (x-4)^2+(y-9)^2=4.$$

So the centre is $C=(4,9)$ and the radius $r=2$.

Step 2. Verify point P and tangent lengths.

Point $P=(2,5)$. Its distance from $C$ is

$$CP=\sqrt{(2-4)^2+(5-9)^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}.$$

Since $2\sqrt{5} > 2$, $P$ is outside. The length of each tangent from $P$ is

$$PQ = \sqrt{CP^2-r^2}=\sqrt{20-4}=\sqrt{16}=4.$$

Step 3. Option (A): Angle between the tangents

The angle between the two tangents from an external point is given by

$$\theta = 2\sin^{-1}\left(\frac{r}{CP}\right)=2\sin^{-1}\left(\frac{2}{2\sqrt5}\right)=2\sin^{-1}\left(\frac{1}{\sqrt5}\right).$$

Alternatively, using the half-angle tangent:

$$\tan\frac{\theta}{2}=\frac{r}{\sqrt{CP^2-r^2}}=\frac{2}{4}=\frac{1}{2} \quad\Longrightarrow\quad \frac{\theta}{2}=\tan^{-1}\frac{1}{2},$$

so

$$\theta = 2\tan^{-1}\frac{1}{2}.$$

But note the trigonometric identity:

$$2\tan^{-1}\frac{1}{2}=\tan^{-1}\frac{4}{3},$$

since $\tan(2\tan^{-1}(1/2))=\frac{2(1/2)}{1-(1/2)^2}=\frac{1}{1-1/4}=\frac{1}{3/4}=\frac{4}{3}$.
Thus (A) is true.

Step 4. Option (B): CP is the arithmetic mean of AP and BP

The line joining $P$ and $C$ meets the circle at $A$ and $B$. Write the line $PC$: through $P(2,5)$ and $C(4,9)$ the slope is

$$m=\frac{9-5}{4-2}=2,$$

so its equation is

$$y-5=2(x-2) \quad\Longrightarrow\quad y=2x+1.$$

Parameterize:

$$( x, y ) = (2+2t,\; 5+4t), \quad \text{with } t=0 \text{ at } P \text{ and } t=1 \text{ at } C.$$

Substitute into the circle’s equation (in centre–radius form):

$$(2+2t-4)^2+(5+4t-9)^2=4 \quad\Longrightarrow\quad (2t-2)^2+(4t-4)^2=4.$$

Factor:

$$4(t-1)^2+16(t-1)^2 =20(t-1)^2=4 \quad\Longrightarrow\quad (t-1)^2=\frac{1}{5}.$$

So,

$$t=1\pm \frac{1}{\sqrt5}.$$

Thus, the distances along $PC$ are:

$$PA = (1-\frac{1}{\sqrt5})\cdot CP,\qquad PB = (1+\frac{1}{\sqrt5})\cdot CP.$$

The arithmetic mean is:

$$\frac{PA+PB}{2}=\frac{\left(1-\frac{1}{\sqrt5}+1+\frac{1}{\sqrt5}\right)CP}{2}=\frac{2CP}{2}=CP.$$

So (B) is true.

Step 5. Option (C): Area of $\triangle PQR$

In $\triangle PQR$ the two tangents $PQ$ and $PR$ (each of length 4) make an angle $\theta$ at $P$ with $\sin\theta$ computed as

$$\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}.$$

Since

$$\sin\frac{\theta}{2}=\frac{r}{CP}=\frac{2}{2\sqrt5}=\frac{1}{\sqrt5},\quad \cos\frac{\theta}{2}=\sqrt{1-\frac{1}{5}}=\frac{2}{\sqrt5},$$

we have:

$$\sin\theta=2\cdot \frac{1}{\sqrt5}\cdot \frac{2}{\sqrt5}=\frac{4}{5}.$$

Thus the area is:

$$\text{Area}=\frac{1}{2}\cdot PQ\cdot PR\cdot \sin\theta= \frac{1}{2}\cdot 4\cdot 4\cdot \frac{4}{5}=\frac{32}{5}.$$

So (C) is true.

Step 6. Option (D): Equation of the circumcircle of $\triangle PQR$

A well‐known result is that for the pair of tangents drawn from an external point $P$ to a circle, the circumcenter of $\triangle PQR$ is the midpoint of $PC$. We have:

$$P=(2,5),\quad C=(4,9)\quad\Longrightarrow\quad \text{Midpoint }O=\left(\frac{2+4}{2},\frac{5+9}{2}\right)=(3,7).$$

Distance $OP$ is:

$$OP=\sqrt{(3-2)^2+(7-5)^2}=\sqrt{1+4}=\sqrt5.$$

Thus the circumcircle of $\triangle PQR$ is

$$(x-3)^2+(y-7)^2=5,$$

which expands to:

$$x^2+y^2-6x-14y+9+49-5=0 \quad\Longrightarrow\quad x^2+y^2-6x-14y+53=0.$$

So (D) is true.

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Summary (Minimal Explanation)

1. Centre & Radius: Completing the square gives $C=(4,9)$, $r=2$.
2. Tangents: $CP=2\sqrt5$ and tangent length $=4$. Using $\tan\frac{\theta}{2}=\frac{1}{2}$, we get angle $2\tan^{-1}(1/2)=\tan^{-1}(4/3)$.
3. Line PC intersections: Parameterizing shows that $CP$ equals the arithmetic mean of $PA$ and $PB$.
4. Area: Using $\sin\theta=\frac{4}{5}$ gives area $\frac{32}{5}$.
5. Circumcircle: The circumcenter is the midpoint of $PC$ i.e. $(3,7)$ and radius $\sqrt5$, leading to $x^2+y^2-6x-14y+53=0$.

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