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Animated Solution for Mathematics - Circles: Equation of two diameters of a circle are $2 x - 3 y = 5$ and $3 x - 4 y = 7$ . The line joining the points $(- \frac{22}{7}, - 4)$ and $(- \frac{1}{7}, 3)$ intersects the circle at only one point $P (α, β)$ . Then $17 β - α$ is equal to

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: The Spark: Visualize the geometry. A circle is defined by its center and radius. The diameters are the skeleton of the circle; they intersect at a single, crucial point. Can you solve the system of linear equations

2 x - 3 y = 5

and

3 x - 4 y = 7

to find the center

C (h, k)

Hint 2

Hint 2: The Arsenal: The problem mentions a line intersecting the circle at exactly one point. In the language of geometry, this is a tangent. Recall the fundamental property: the radius drawn to the point of contact is perpendicular to the tangent. You need to find the coordinates of this point of contact

P (α, β)

Hint 3

Hint 3: The Execution: To find the coordinates of the point of contact

P

, you are looking for the 'foot of the perpendicular' from the center

C

to the tangent line. Use the formula:

\frac{α - x _{1}}{a} = \frac{β - y _{1}}{b} = - \frac{a x _{1} + b y _{1} + c}{a ^{2} + b ^{2}}

, where

(x_{1}, y_{1})

is the center

C

and

a x + b y + c = 0

is the equation of the tangent.

Visualized Solution

Visualizing the Diameters

Diameters of the circle:
Line 1: $2 x - 3 y = 5$
Line 2: $3 x - 4 y = 7$
Center $C (h, k)$ is the intersection of these diameters.

Solving for the Center $C$

Multiply Line 1 by 3: $6 x - 9 y = 15$
Multiply Line 2 by 2: $6 x - 8 y = 14$
Subtracting gives: $- y = 1 \Rightarrow y = - 1$
Substitute $y = - 1$ into Line 1: $2 x - 3 (- 1) = 5 \Rightarrow x = 1$
Center: $C = (1, - 1)$

The Tangent Line $A B$

The line joins points $A (- \frac{22}{7}, - 4)$ and $B (- \frac{1}{7}, 3)$ .
It intersects the circle at only one point, so it must be a tangent.
Slope of $A B$ ( $m$ ):
$m = \frac{3 - ( - 4 )}{- \frac{1}{7} - ( - \frac{22}{7} )} = \frac{7}{\frac{21}{7}} = \frac{7}{3}$

Equation of the Tangent

Using point-slope form with $B (- \frac{1}{7}, 3)$ :
$y - 3 = \frac{7}{3} (x + \frac{1}{7})$
$3 y - 9 = 7 x + 1$
Tangent Equation: $7 x - 3 y + 10 = 0$

Circle and Point of Contact $P$

The tangent touches the circle at exactly one point $P (α, β)$ .
The radius from the center $C (1, - 1)$ to $P$ is perpendicular to the tangent.
Therefore, $P$ is the foot of the perpendicular from $C$ to the tangent line.

Foot of Perpendicular Formula

Formula for foot of perpendicular $(x_{2}, y_{2})$ from $(x_{1}, y_{1})$ to $a x + b y + c = 0$ :
$\frac{x _{2} - x _{1}}{a} = \frac{y _{2} - y _{1}}{b} = - \frac{a x _{1} + b y _{1} + c}{a ^{2} + b ^{2}}$
Substituting our values:
$\frac{α - 1}{7} = \frac{β - ( - 1 )}{- 3} = - \frac{7 ( 1 ) - 3 ( - 1 ) + 10}{7 ^{2} + ( - 3 ) ^{2}}$

Simplifying the Ratio

Evaluate the right-hand side:
Numerator: $7 (1) + 3 + 10 = 20$
Denominator: $49 + 9 = 58$
Ratio: - $\frac{20}{58} = - \frac{10}{29}$
So, $\frac{α - 1}{7} = \frac{β + 1}{- 3} = - \frac{10}{29}$

Solving for $α$

Equate the $α$ term to the ratio:
$\frac{α - 1}{7} = - \frac{10}{29}$
$α - 1 = - \frac{70}{29}$
$α = 1 - \frac{70}{29} = - \frac{41}{29}$

Solving for $β$

Equate the $β$ term to the ratio:
$\frac{β + 1}{- 3} = - \frac{10}{29}$
$β + 1 = \frac{30}{29}$
$β = \frac{30}{29} - 1 = \frac{1}{29}$

Final Result: $17 β - α$

We need to find the value of $17 β - α$ .
Substitute $α = - \frac{41}{29}$ and $β = \frac{1}{29}$ :
$17 (\frac{1}{29}) - (- \frac{41}{29})$
$= \frac{17 + 41}{29} = \frac{58}{29} = 2$
Final Answer: $2$

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

To find the center of the circle, we recognize that the intersection of two diameters is the center

C (h, k)

. We are given the equations:

2 x - 3 y = 5

3 x - 4 y = 7

Using the elimination method, we multiply the first equation by

3

and the second by

2

6 x - 9 y = 15

6 x - 8 y = 14

Subtracting the first from the second yields

y = - 1

. Substituting

y = - 1

into the first equation, we find

x = 1

. Thus, the center of the circle is

C (1, - 1)

The Tangent

A Line of Precision

The tangent line passes through the points

(- \frac{22}{7}, - 4)

and

(- \frac{1}{7}, 3)

. We calculate the slope

m

using the formula:

m = \frac{3 - ( - 4 )}{- \frac{1}{7} - ( - \frac{22}{7} )} = \frac{7}{3}

Using the point-slope form

y - y_{1} = m (x - x_{1})

with the point

(- \frac{1}{7}, 3)

, we have:

y - 3 = \frac{7}{3} (x + \frac{1}{7})

Multiplying by

3

and simplifying, we obtain:

3 y - 9 = 7 x + 1 \Rightarrow 7 x - 3 y + 10 = 0

The Foot of the Perpendicular

The Point of Contact

The point of contact

P (α, β)

is the foot of the perpendicular from the center

C (1, - 1)

to the tangent line

7 x - 3 y + 10 = 0

. We apply the perpendicular foot formula:

\frac{α - 1}{7} = \frac{β - ( - 1 )}{- 3} = - \frac{7 ( 1 ) - 3 ( - 1 ) + 10}{7 ^{2} + ( - 3 ) ^{2}}

Evaluating the constant term:

- \frac{7 + 3 + 10}{49 + 9} = - \frac{20}{58} = - \frac{10}{29}

The Final Resolution

We now solve for the coordinates

α

and

β

individually:

α - 1 = 7 \times (- \frac{10}{29}) = - \frac{70}{29} \Rightarrow α = 1 - \frac{70}{29} = - \frac{41}{29}

β + 1 = - 3 \times (- \frac{10}{29}) = \frac{30}{29} \Rightarrow β = \frac{30}{29} - 1 = \frac{1}{29}

Finally, we compute the required expression

17 β - α

17 (\frac{1}{29}) - (- \frac{41}{29}) = \frac{17 + 41}{29} = \frac{58}{29} = 2

The final result is 2.

Common Traps & FAQs

Wait, how do I know for sure that the line is a tangent?

That is a fantastic question! The problem explicitly states that the line intersects the circle at 'only one point'. By definition, a line that touches a circle at exactly one point is a tangent. It is a geometric guarantee, not an assumption!

Why can't I just use the distance formula to find the point of contact?

The distance formula is excellent for finding the length of a segment, but it won't give you the coordinates

(α, β)

directly. Since the question asks for a specific combination of

α

and

β

, we need the exact coordinates. The 'foot of the perpendicular' method is the most direct and elegant way to find the point of contact.

Is there a trick to solving the system of equations for the center?

No tricks, just steady, methodical elimination! When you have two linear equations, the goal is to eliminate one variable. Multiply the equations by constants to make the coefficients of

x

y

match, then subtract. Keep your signs clean—that is where most students lose points!

Animated Solution for Mathematics - Circles: Equation of two diameters of a circle are $2 x - 3 y = 5$ and $3 x - 4 y = 7$ . The line joining the points $(- \frac{22}{7}, - 4)$ and $(- \frac{1}{7}, 3)$ intersects the circle at only one point $P (α, β)$ . Then $17 β - α$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Diameters

Solving for the Center $C$

The Tangent Line $A B$

Equation of the Tangent

Circle and Point of Contact $P$

Foot of Perpendicular Formula

Simplifying the Ratio

Solving for $α$

Solving for $β$

Final Result: $17 β - α$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent

The Foot of the Perpendicular

The Final Resolution

Similar Questions

List-I

List-II

Two circles each of radius 5 units touch each other at the point $(1, 2)$ . If the equation of their common tangent is $4 x + 3 y = 10$ , and $C_{1} (α, β)$ and $C_{2} (γ, δ)$ , $C_{1} \neq = C_{2}$ are their centres, then $∣ (α + β) (γ + δ) ∣$ is equal to .

Let the centre of a circle $C$ be $(α, β)$ and its radius $r < 8$ . Let $3 x + 4 y = 24$ and $3 x - 4 y = 32$ be two tangents and $4 x + 3 y = 1$ be a normal to $C$ . Then $(α - β + r)$ is equal to

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Let the equation of two diameters of a circle $x^{2} + y^{2} - 2 x + 2 f y + 1 = 0$ be $2 p x - y = 1$ and $2 x + p y = 4 p$ . Then the slope $m \in (0, \infty)$ of the tangent to the hyperbola $3 x^{2} - y^{2} = 3$ passing through the centre of the circle is equal to ______.

If a line, $y = m x + c$ is a tangent to the circle, $(x - 3)^{2} + y^{2} = 1$ and it is perpendicular to a line $L_{1}$ , where $L_{1}$ is the tangent to the circle, $x^{2} + y^{2} = 1$ at the point $(\frac{1}{2}, \frac{1}{2})$ :

Let $P Q$ and $R S$ be tangents at the extremities of the diameter $P R$ of a circle of radius $r$ . If $P S$ and $R Q$ intersect at a point $X$ on the circumference of the circle, then $2 r$ equals

The circle $x^{2} + y^{2} = 4 x + 8 y + 5$ intersects the line $3 x - 4 y = m$ at two distinct points if

Enter Numerical Value:

Visualized Solution

Visualizing the Diameters

Solving for the Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

The Tangent Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Equation of the Tangent

Circle and Point of Contact .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Foot of Perpendicular Formula

Simplifying the Ratio

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Final Result: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 17β−α

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent

The Foot of the Perpendicular

The Final Resolution

Similar Questions

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the centre of a circle C be (α,β) and its radius r<8. Let 3x+4y=24 and 3x−4y=32 be two tangents and 4x+3y=1 be a normal to C. Then (α−β+r) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The circle x2+y2=4x+8y+5 intersects the line 3x−4y=m at two distinct points if

Solving for the Center $C$

The Tangent Line $A B$

Circle and Point of Contact $P$

Solving for $α$

Solving for $β$

Final Result: $17 β - α$

Let the centre of a circle $C$ be $(α, β)$ and its radius $r < 8$ . Let $3 x + 4 y = 24$ and $3 x - 4 y = 32$ be two tangents and $4 x + 3 y = 1$ be a normal to $C$ . Then $(α - β + r)$ is equal to

The circle $x^{2} + y^{2} = 4 x + 8 y + 5$ intersects the line $3 x - 4 y = m$ at two distinct points if