Wait, why is the length of the tangent simply $\sqrt{S_1}$? Can't we use the distance formula?

You absolutely can! If you find the center of the circle $C(-g, -f)$ and the radius $r$, you can form a right-angled triangle $\triangle PQC$ where $\angle QPC = 90^\circ$. By Pythagoras, $PQ^2 + r^2 = QC^2$, so $PQ = \sqrt{QC^2 - r^2}$. The formula $\sqrt{S_1}$ is just a beautiful, condensed algebraic shortcut for this exact geometric relationship. It saves you from calculating the center and radius explicitly!

LEVELJEE Main

Animated Solution for Mathematics - Circles: If the tangent at the point $P$ on the circle $x^{2} + y^{2} + 6 x + 6 y = 2$ meets a straight line $5 x - 2 y + 6 = 0$ at a point $Q$ on the y-axis, then the length of $P Q$ is

Q: What happens if the point $Q$ is inside the circle? Does the formula still work?

Great question! If $Q$ is inside the circle, $S_1$ will be negative. Since we cannot take the square root of a negative number in the real plane, this confirms that no real tangent can be drawn from an interior point to the circle. The formula $\sqrt{S_1}$ naturally alerts you to this by becoming undefined in the real number system.

Q: Is it possible for the tangent to be vertical or horizontal?

Yes, absolutely. The formula $\sqrt{S_1}$ is independent of the orientation of the tangent. It relies solely on the power of the point with respect to the circle, which is a scalar property of the point's position relative to the circle's boundary. It works regardless of the slope of the tangent line.

Select Answer:

Visualized Solution

Visualize the Geometry

Circle Equation: $x^{2} + y^{2} + 6 x + 6 y - 2 = 0$
Line Equation: $5 x - 2 y + 6 = 0$
We need the length of tangent $P Q$ from point $Q$ on the line to the circle.

Setup to Find Point $Q$

Point $Q$ lies on the y-axis, so its x-coordinate is $0$ .
Let $Q = (0, y)$ .
Substitute $x = 0$ into the line equation:
$5 (0) - 2 y + 6 = 0$

Calculate Coordinates of $Q$

Simplify the equation: $- 2 y + 6 = 0$
Transpose $6$ : $- 2 y = - 6$
Divide by $- 2$ : $y = 3$
Therefore, point $Q = (0, 3)$

Visualize the Tangent $P Q$

From $Q (0, 3)$ , a tangent is drawn to the circle.
Let the point of tangency be $P$ .
We need to find the length of the segment $P Q$ .

The Tangent Length Formula

The length of a tangent from an external point $(x_{1}, y_{1})$ to a circle $S = 0$ is:
$L = S_{1}$
Where $S_{1} = x_{1}^{2} + y_{1}^{2} + 6 x_{1} + 6 y_{1} - 2$

Substitute $Q$ into the Circle Equation

Substitute $Q (0, 3)$ into the expression for $S_{1}$ :
$S_{1} = (0)^{2} + (3)^{2} + 6 (0) + 6 (3) - 2$
This represents the power of point $Q$ with respect to the circle.

Compute the Power of Point ( $S_{1}$ )

Calculate the individual terms:
$S_{1} = 0 + 9 + 0 + 18 - 2$
Summing the values:
$S_{1} = 27 - 2 = 25$

Final Result for Length $P Q$

Substitute $S_{1} = 25$ into the length formula:
$P Q = 25$
$P Q = 5$
The length of the tangent is $5$ units.

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The Sigma Insight: Length of Tangent and Chord of Contact

Solution Diagram

Analyzing the Setup

The circle is defined by the equation:

x^{2} + y^{2} + 6 x + 6 y - 2 = 0

The boundary line is given by:

5 x - 2 y + 6 = 0

We are tasked with finding the length of the tangent

P Q

drawn from a point

Q

to the circle, where

Q

lies on both the y-axis and the given line.

Locating the Mysterious Point $Q$

Since point

Q

lies on the y-axis, its x-coordinate must be

0

. We substitute

x = 0

into the equation of the line

5 x - 2 y + 6 = 0

to find the corresponding y-coordinate:

5 (0) - 2 y + 6 = 0

- 2 y = - 6

y = 3

Thus, our point of departure is $Q (0, 3)$ .

The Power of a Point

To find the length of the tangent

P Q

, we utilize the Power of a Point theorem. For a circle defined by

S : x^{2} + y^{2} + 2 g x + 2 f y + c = 0

, the length of the tangent

L

from an external point

(x_{1}, y_{1})

is given by:

L = S_{1} = x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c

This theorem allows us to bypass complex geometric constructions by directly evaluating the circle's equation at the coordinates of point

Q

The Final Calculation

We substitute the coordinates of

Q (0, 3)

into the circle's equation

S = x^{2} + y^{2} + 6 x + 6 y - 2

S_{1} = (0)^{2} + (3)^{2} + 6 (0) + 6 (3) - 2

Calculating the terms individually:

S_{1} = 0 + 9 + 0 + 18 - 2

S_{1} = 27 - 2 = 25

The length of the tangent

P Q

is the square root of the power of the point:

P Q = 25 = 5

The final length of the tangent segment

P Q

is $5$ .

Common Traps & FAQs

Wait, why is the length of the tangent simply

S_{1}

? Can't we use the distance formula?

You absolutely can! If you find the center of the circle

C (- g, - f)

and the radius

r

, you can form a right-angled triangle

△ P QC

where

∠ QP C = 9 0^{\circ}

. By Pythagoras,

P Q^{2} + r^{2} = Q C^{2}

, so

P Q = Q C^{2} - r^{2}

. The formula

S_{1}

is just a beautiful, condensed algebraic shortcut for this exact geometric relationship. It saves you from calculating the center and radius explicitly!

What happens if the point

Q

is inside the circle? Does the formula still work?

Great question! If

Q

is inside the circle,

S_{1}

will be negative. Since we cannot take the square root of a negative number in the real plane, this confirms that no real tangent can be drawn from an interior point to the circle. The formula

S_{1}

naturally alerts you to this by becoming undefined in the real number system.

Is it possible for the tangent to be vertical or horizontal?

Yes, absolutely. The formula

S_{1}

is independent of the orientation of the tangent. It relies solely on the power of the point with respect to the circle, which is a scalar property of the point's position relative to the circle's boundary. It works regardless of the slope of the tangent line.

Animated Solution for Mathematics - Circles: If the tangent at the point $P$ on the circle $x^{2} + y^{2} + 6 x + 6 y = 2$ meets a straight line $5 x - 2 y + 6 = 0$ at a point $Q$ on the y-axis, then the length of $P Q$ is

Select Answer:

Visualized Solution

Visualize the Geometry

Setup to Find Point $Q$

Calculate Coordinates of $Q$

Visualize the Tangent $P Q$

The Tangent Length Formula

Substitute $Q$ into the Circle Equation

Compute the Power of Point ( $S_{1}$ )

Final Result for Length $P Q$

The Sigma Insight: Length of Tangent and Chord of Contact

Analyzing the Setup

Locating the Mysterious Point $Q$

The Power of a Point

The Final Calculation

Similar Questions

If the tangents at the points $P$ and $Q$ on the circle $x^{2} + y^{2} - 2 x + y = 5$ meet at the point $R (\frac{9}{4}, 2)$ , then the area of the triangle $P QR$ is

A circle with centre $(2, 3)$ and radius 4 intersects the line $x + y = 3$ at the points $P$ and $Q$ . If the tangents at $P$ and $Q$ intersect at the point $S (α, β)$ , then $4 α - 7 β$ is equal to

Let $B$ be the centre of the circle $x^{2} + y^{2} - 2 x + 4 y + 1 = 0$ . Let the tangents at two points $P$ and $Q$ on the circle intersect at the point $A (3, 1)$ . Then $8 \cdot (\frac{area Δ A P Q}{area Δ B P Q})$ is equal to .

Let the tangents drawn from the origin to the circle, $x^{2} + y^{2} - 8 x - 4 y + 16 = 0$ touch it at the point $A$ and $B$ . The $(A B)^{2}$ is equal to:

The area of the triangle formed by the tangents from the point $(4, 3)$ to the the circle $x^{2} + y^{2} = 9$ and the line joining their points of contact is .........

Comprehension Passage

Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curve

The chords of contact of the pair of tangents drawn from each point on the line $2 x + y = 4$ to circle $x^{2} + y^{2} = 1$ pass through the point .........

Let the tangents drawn from the origin to the circle, $x^{2} + y^{2} - 8 x - 4 y + 16 = 0$ touch it at the points $A$ and $B$ . The $(A B)^{2}$ is equal to :

For the circle $x^{2} + y^{2} = r^{2}$ , find the value of $r$ for which the area enclosed by the tangents drawn from the point $P (6, 8)$ to the circle and the chord of contact is maximum.

Select Answer:

Visualized Solution

Visualize the Geometry

Setup to Find Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Calculate Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Visualize the Tangent .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ

The Tangent Length Formula

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q into the Circle Equation

Compute the Power of Point ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​)

Final Result for Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ

The Sigma Insight: Length of Tangent and Chord of Contact

Analyzing the Setup

Locating the Mysterious Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

The Power of a Point

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangents at the points P and Q on the circle x2+y2−2x+y=5 meet at the point R(49​,2), then the area of the triangle PQR is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the tangents drawn from the origin to the circle, x2+y2−8x−4y+16=0 touch it at the point A and B. The (AB)2 is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the triangle formed by the tangents from the point (4,3) to the the circle x2+y2=9 and the line joining their points of contact is .........

Comprehension Passage

Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curve

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The chords of contact of the pair of tangents drawn from each point on the line 2x+y=4 to circle x2+y2=1 pass through the point .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the tangents drawn from the origin to the circle, x2+y2−8x−4y+16=0 touch it at the points A and B. The (AB)2 is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For the circle x2+y2=r2, find the value of r for which the area enclosed by the tangents drawn from the point P(6,8) to the circle and the chord of contact is maximum.

Setup to Find Point $Q$

Calculate Coordinates of $Q$

Visualize the Tangent $P Q$

Substitute $Q$ into the Circle Equation

Compute the Power of Point ( $S_{1}$ )

Final Result for Length $P Q$

Locating the Mysterious Point $Q$

If the tangents at the points $P$ and $Q$ on the circle $x^{2} + y^{2} - 2 x + y = 5$ meet at the point $R (\frac{9}{4}, 2)$ , then the area of the triangle $P QR$ is

Let the tangents drawn from the origin to the circle, $x^{2} + y^{2} - 8 x - 4 y + 16 = 0$ touch it at the point $A$ and $B$ . The $(A B)^{2}$ is equal to:

The area of the triangle formed by the tangents from the point $(4, 3)$ to the the circle $x^{2} + y^{2} = 9$ and the line joining their points of contact is .........

The chords of contact of the pair of tangents drawn from each point on the line $2 x + y = 4$ to circle $x^{2} + y^{2} = 1$ pass through the point .........

Let the tangents drawn from the origin to the circle, $x^{2} + y^{2} - 8 x - 4 y + 16 = 0$ touch it at the points $A$ and $B$ . The $(A B)^{2}$ is equal to :

For the circle $x^{2} + y^{2} = r^{2}$ , find the value of $r$ for which the area enclosed by the tangents drawn from the point $P (6, 8)$ to the circle and the chord of contact is maximum.