Home
Class 12
MATHS
The lines 2x-5y+1=0 and 10x-4y-3=0 meets...

The lines `2x-5y+1=0` and `10x-4y-3=0` meets the coordinate axes in concylic points, then equation to the circle is

A

`20x^(2)+20y^(2)-4x-11y=0`

B

`20x^(2)+20y^(2)+4x-11y-3=0`

C

`20x^(2)+20y^(2)+4x+11y-3=0`

D

None

Text Solution

AI Generated Solution

The correct Answer is:
To find the equation of the circle that passes through the points where the lines \(2x - 5y + 1 = 0\) and \(10x - 4y - 3 = 0\) meet the coordinate axes, we will follow these steps: ### Step 1: Find the points where the first line intersects the axes. 1. **Finding the y-intercept** (set \(x = 0\)): \[ 2(0) - 5y + 1 = 0 \implies -5y + 1 = 0 \implies y = \frac{1}{5} \] So, the point is \(A(0, \frac{1}{5})\). 2. **Finding the x-intercept** (set \(y = 0\)): \[ 2x - 5(0) + 1 = 0 \implies 2x + 1 = 0 \implies x = -\frac{1}{2} \] So, the point is \(B(-\frac{1}{2}, 0)\). ### Step 2: Find the points where the second line intersects the axes. 1. **Finding the y-intercept** (set \(x = 0\)): \[ 10(0) - 4y - 3 = 0 \implies -4y - 3 = 0 \implies y = -\frac{3}{4} \] So, the point is \(C(0, -\frac{3}{4})\). 2. **Finding the x-intercept** (set \(y = 0\)): \[ 10x - 4(0) - 3 = 0 \implies 10x - 3 = 0 \implies x = \frac{3}{10} \] So, the point is \(D(\frac{3}{10}, 0)\). ### Step 3: List the points. The points we have found are: - \(A(0, \frac{1}{5})\) - \(B(-\frac{1}{2}, 0)\) - \(C(0, -\frac{3}{4})\) - \(D(\frac{3}{10}, 0)\) ### Step 4: Set up the equation of the circle. The general equation of a circle is: \[ x^2 + y^2 + 2gx + 2fy + c = 0 \] ### Step 5: Substitute the points into the circle equation. 1. **Using point A(0, \frac{1}{5})**: \[ 0^2 + \left(\frac{1}{5}\right)^2 + 2g(0) + 2f\left(\frac{1}{5}\right) + c = 0 \] \[ \frac{1}{25} + \frac{2f}{5} + c = 0 \quad \text{(Equation 1)} \] 2. **Using point B(-\frac{1}{2}, 0)**: \[ \left(-\frac{1}{2}\right)^2 + 0^2 + 2g\left(-\frac{1}{2}\right) + 2f(0) + c = 0 \] \[ \frac{1}{4} - g + c = 0 \quad \text{(Equation 2)} \] 3. **Using point C(0, -\frac{3}{4})**: \[ 0^2 + \left(-\frac{3}{4}\right)^2 + 2g(0) + 2f\left(-\frac{3}{4}\right) + c = 0 \] \[ \frac{9}{16} - \frac{3f}{2} + c = 0 \quad \text{(Equation 3)} \] 4. **Using point D(\frac{3}{10}, 0)**: \[ \left(\frac{3}{10}\right)^2 + 0^2 + 2g\left(\frac{3}{10}\right) + 2f(0) + c = 0 \] \[ \frac{9}{100} + \frac{3g}{5} + c = 0 \quad \text{(Equation 4)} \] ### Step 6: Solve the equations. We will solve these equations step by step to find \(g\), \(f\), and \(c\). 1. From Equation 1: \[ c = -\frac{1}{25} - \frac{2f}{5} \] 2. Substitute \(c\) from Equation 1 into Equation 2: \[ \frac{1}{4} - g - \left(-\frac{1}{25} - \frac{2f}{5}\right) = 0 \] Simplifying gives: \[ \frac{1}{4} - g + \frac{1}{25} + \frac{2f}{5} = 0 \] 3. Substitute \(c\) from Equation 1 into Equation 3: \[ \frac{9}{16} - \frac{3f}{2} - \left(-\frac{1}{25} - \frac{2f}{5}\right) = 0 \] Simplifying gives: \[ \frac{9}{16} + \frac{1}{25} - \frac{3f}{2} - \frac{2f}{5} = 0 \] 4. Continue solving these equations to find the values of \(g\), \(f\), and \(c\). ### Final Step: Write the equation of the circle. Once \(g\), \(f\), and \(c\) are found, substitute them back into the general equation of the circle to get the final equation.
Promotional Banner

Similar Questions

Explore conceptually related problems

Show that the lines 3x-y+3=0 and x-3y-6=0 cut the coordinate axes at concylic points. Show that the equation of the circle passing through these points is x^(2)+y^(2)-5x-y-6=0

If the lines lx+2y +3=0 and 2y+mx+4=0 cut the co-ordinate axes in concylic points then lm=

The lines 2x+3y+19=0 and 9x+6y-17=0 , cut the coordinate axes at concyclic points.

The lines 2x+3y+19=0 and 9x+6y-17=0 cuts the coordinate axes in

If the lines x-2y+3=0, 3x+ky+7=0 cut the coordinate axes in concylic points, then k=

If the lines 2x-3y+7=0, 3x+ky+5=0 cut the coordinate axes in concyclic points then k=

If the 4 points made by intersection of lines 2x-y+1=0, x-2y+3=0 with the coordinate axes are concylic then centre of circle is

If the lines 2x+3y+1=0, 6x+4y+1=0 intersect the co-ordinate axes in 4 points, then the circle passing through the points is

The algebraic sum of distances of the line ax + by + 2 = 0 from (1,2), (2,1) and (3,5) is zero and the lines bx - ay + 4 = 0 and 3x + 4y + 5=0 cut the coordinate axes at concyclic points. Then (a) a+b=-2/7 (b) area of triangle formed by the line ax+by+2=0 with coordinate axes is 14/5 (c) line ax+by+3=0 always passes through the point (-1,1) (d) max {a,b}=5/7

A line passes through (-3,4) and the portion of the line intercepted between the coordinate axes is bisected at the point then equation of line is (A) 4x-3y+24=0 (B) x-y-7=0 (C) 3x-4y+25=0 (D) 3x-4y+24=0