Equation of two diameters of a circle are \(2x-3y=5\) and \(3x-4y=7\).The line joining the points \((-\frac{22}{7},-4)\) and \((-\frac{1}{7},3)\) intersects the circle at only one point \(P(\alpha,\beta)\).Then \(17\beta-\alpha\) is equal to.
Correct Answer: 2
Approach Solution - 1
Step 1: Find the center of the circle.
The diameters intersect at the center of the circle. So, we solve the two linear equations: \[ 2x - 3y = 5 \quad \text{and} \quad 3x - 4y = 7 \]
Multiply the first equation by 3 and the second by 2 to eliminate \( x \): \[ 6x - 9y = 15 \\ 6x - 8y = 14 \] Subtracting: \[ (6x - 9y) - (6x - 8y) = 15 - 14 \] \[ -y = 1 \implies y = -1 \]
Substitute \( y = -1 \) in \( 2x - 3y = 5 \):
\[ 2x - 3(-1) = 5 \implies 2x + 3 = 5 \implies 2x = 2 \implies x = 1 \]
Therefore, the center of the circle is: \[ C(1, -1) \]
Step 2: Equation of the line joining given points
Let the two points be: \[ A\left(-\frac{22}{7}, -4\right) \quad \text{and} \quad B\left(-\frac{1}{7}, 3\right) \]
Slope of line \( AB \): \[ m = \frac{3 - (-4)}{-\frac{1}{7} - (-\frac{22}{7})} = \frac{7}{\frac{21}{7}} = \frac{7}{3} \]
Equation of line using point \( A(x_1, y_1) \): \[ y + 4 = \frac{7}{3}\left(x + \frac{22}{7}\right) \]
Simplify: \[ 3(y + 4) = 7\left(x + \frac{22}{7}\right) \] \[ 3y + 12 = 7x + 22 \] \[ 7x - 3y + 10 = 0 \]
Hence, the equation of the line is: \[ 7x - 3y + 10 = 0 \]
Step 3: Condition for line to touch the circle (one point)
The line \( 7x - 3y + 10 = 0 \) is tangent to the circle. The perpendicular distance from the center \( (1, -1) \) to this line equals the radius \( r \).
Distance from \( (x_1, y_1) \) to \( ax + by + c = 0 \): \[ d = \frac{|a x_1 + b y_1 + c|}{\sqrt{a^2 + b^2}} \]
Substitute \( a = 7, b = -3, c = 10, (x_1, y_1) = (1, -1) \): \[ d = \frac{|7(1) - 3(-1) + 10|}{\sqrt{7^2 + (-3)^2}} = \frac{|7 + 3 + 10|}{\sqrt{49 + 9}} = \frac{20}{\sqrt{58}} \] \[ r = \frac{20}{\sqrt{58}} \]
Step 4: Equation of the circle
\[ (x - 1)^2 + (y + 1)^2 = r^2 = \frac{400}{58} = \frac{200}{29} \] \] \[ (x - 1)^2 + (y + 1)^2 = \frac{200}{29} \]
Step 5: Find the point of tangency (P)
Equation of circle: \[ (x - 1)^2 + (y + 1)^2 = \frac{200}{29} \] Equation of tangent line: \[ 7x - 3y + 10 = 0 \] \[ \Rightarrow y = \frac{7x + 10}{3} \]
Substitute in circle equation: \[ (x - 1)^2 + \left(\frac{7x + 10}{3} + 1\right)^2 = \frac{200}{29} \] \[ (x - 1)^2 + \left(\frac{7x + 13}{3}\right)^2 = \frac{200}{29} \]
Simplify: \[ 9(x - 1)^2 + (7x + 13)^2 = \frac{1800}{29} \] \[ 9(x^2 - 2x + 1) + (49x^2 + 182x + 169) = \frac{1800}{29} \] \[ 58x^2 + 164x + 178 = \frac{1800}{29} \] \[ \Rightarrow 1682x^2 + 4756x + 5162 - 1800 = 0 \] \[ 1682x^2 + 4756x + 3362 = 0 \] \] \[ \text{Since line is tangent, discriminant = 0:} \] \[ (4756)^2 - 4(1682)(3362) = 0 \] \] Thus one point of contact \( P(\alpha, \beta) \).
Solve for \( x \): \[ \alpha = -\frac{b}{2a} = -\frac{4756}{2(1682)} = -\frac{4756}{3364} = -\frac{1189}{841} \]
Now \( y = \frac{7x + 10}{3} \): \[ \beta = \frac{7\left(-\frac{1189}{841}\right) + 10}{3} = \frac{-8323 + 8410}{2523} = \frac{87}{2523} = \frac{29}{841} \]
Step 6: Find \( 17\beta - \alpha \)
\[ 17\beta - \alpha = 17\left(\frac{29}{841}\right) - \left(-\frac{1189}{841}\right) \] \[ = \frac{493 + 1189}{841} = \frac{1682}{841} = 2 \]
Final Answer:
\[ \boxed{17\beta - \alpha = 2} \]
Approach Solution -2
Step 1: Find the Centre of the Circle
The centre \(C\) of the circle is the intersection of the diameters \(2x - 3y = 5\) and \(3x - 4y = 7\). Solving these equations, we get \(C(1, -1)\).
Step 2: Equation of Line \(AB\)
The points \(A\left(-\frac{22}{7}, -4\right)\) and \(B\left(\frac{1}{7}, 3\right)\) lie on the line \(AB\). The equation of \(AB\) is:
\[ 7x - 3y + 10 = 0 \quad (i) \]
Step 3: Equation of Line \(CP\)
Since \(P\) lies on the circle, \(CP\) is perpendicular to \(AB\) with the equation:
\[ 3x + 7y + 4 = 0 \quad (ii) \]
Step 4: Solve for \(\alpha\) and \(\beta\)
Solving equations (i) and (ii), we find:
\[ \alpha = -\frac{41}{29}, \quad \beta = \frac{1}{29} \]
Step 5: Calculate \(17\beta - \alpha\)
\[ 17\beta - \alpha = 17 \cdot \frac{1}{29} + \frac{41}{29} = 2 \]
So, the correct answer is: 2
Top Questions on Circles
- Let $ ABC $ be the triangle such that the equations of lines $ AB $ and $ AC $ are: $ 3y - x = 2 \quad \text{and} \quad x + y = 2, $ respectively, and the points $ B $ and $ C $ lie on the x-axis. If $ P $ is the orthocentre of the triangle $ ABC $, then the area of the triangle $ PBC $ is equal to:
- Let $ C_1 $ be the circle in the third quadrant of radius 3, that touches both coordinate axes. Let $ C_2 $ be the circle with center $ (1, 3) $ that touches $ C_1 $ externally at the point $ (\alpha, \beta) $. If $ (\beta - \alpha)^2 = \frac{m}{n} $, and $ \gcd(m, n) = 1 $, then $ m + n $ is equal to:
- The absolute difference between the squares of the radii of the two circles passing through the point \( (-9, 4) \) and touching the lines \( x + y = 3 \) and \( x - y = 3 \), is equal to:
In the following figure chord MN and chord RS intersect at point D. If RD = 15, DS = 4, MD = 8, find DN by completing the following activity:
Activity :
\(\therefore\) MD \(\times\) DN = \(\boxed{\phantom{SD}}\) \(\times\) DS \(\dots\) (Theorem of internal division of chords)
\(\therefore\) \(\boxed{\phantom{8}}\) \(\times\) DN = 15 \(\times\) 4
\(\therefore\) DN = \(\frac{\boxed{\phantom{60}}}{8}\)
\(\therefore\) DN = \(\boxed{\phantom{7.5}}\)In the following figure, circle with centre D touches the sides of \(\angle\)ACB at A and B. If \(\angle\)ACB = 52\(^\circ\), find measure of \(\angle\)ADB.
Questions Asked in JEE Main exam
A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is _____.
- JEE Main - 2025
- Magnetic Field
- Concentrated nitric acid is labelled as 75%by mass. The volume in mL of the solution which contains 30 g of nitric acid is: Given: Density of nitric acid solution is 1.25 g/mL.
- JEE Main - 2025
- Solutions
- A physical quantity \( Q \) is related to four observables \( a \), \( b \), \( c \), and \( d \) as follows: \[ Q = \frac{a b^4}{c d^2} \] Where:
- \( a = (60 \pm 3) \, \text{Pa} \),
- \( b = (20 \pm 0.1) \, \text{m} \),
- \( c = (40 \pm 0.2) \, \text{N·s/m}^2 \),
- \( d = (50 \pm 0.1) \, \text{m} \).
Then the percentage error in \( Q \) is:- JEE Main - 2025
- Error analysis
- Let circle $C$ be the image of
$$ x^2 + y^2 - 2x + 4y - 4 = 0 $$
in the line
$$ 2x - 3y + 5 = 0 $$
and $A$ be the point on $C$ such that $OA$ is parallel to the x-axis and $A$ lies on the right-hand side of the centre $O$ of $C$.
If $B(\alpha, \beta)$, with $\beta < 4$, lies on $C$ such that the length of the arc $AB$ is $\frac{1}{6}$ of the perimeter of $C$, then $\beta - \sqrt{3}\alpha$ is equal to: - Assume a living cell with 0.9%(\(w/w\)) of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data up to first decimal place only) The cell will:
- JEE Main - 2025
- osmosis