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The order of the differential equation o...

The order of the differential equation of family of circles touching two given circles externally is

A

1

B

2

C

3

D

none of these

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The correct Answer is:
To determine the order of the differential equation of the family of circles touching two given circles externally, we can follow these steps: ### Step 1: Define the Circles Let the equations of the two given circles be: 1. Circle 1: \((x - a_1)^2 + (y - b_1)^2 = r_1^2\) 2. Circle 2: \((x - a_2)^2 + (y - b_2)^2 = r_2^2\) ### Step 2: General Equation of the Family of Circles The general equation of a circle can be expressed as: \[ (x - c)^2 + (y - d)^2 = r^2 \] where \(c\), \(d\), and \(r\) are parameters that define the circle. ### Step 3: Set Up the Condition for External Tangency For the family of circles to touch the two given circles externally, the distance between the centers of the circles must equal the sum of their radii. This gives us two equations based on the tangency condition: 1. For Circle 1: \[ \sqrt{(c - a_1)^2 + (d - b_1)^2} = r + r_1 \] 2. For Circle 2: \[ \sqrt{(c - a_2)^2 + (d - b_2)^2} = r + r_2 \] ### Step 4: Square Both Sides to Eliminate the Square Root Squaring both sides of the equations gives: 1. \((c - a_1)^2 + (d - b_1)^2 = (r + r_1)^2\) 2. \((c - a_2)^2 + (d - b_2)^2 = (r + r_2)^2\) ### Step 5: Rearranging the Equations Rearranging the equations leads to: 1. \((c - a_1)^2 + (d - b_1)^2 - (r + r_1)^2 = 0\) 2. \((c - a_2)^2 + (d - b_2)^2 - (r + r_2)^2 = 0\) ### Step 6: Eliminate the Radius \(r\) To find a relationship between \(c\), \(d\), and the parameters of the given circles, we can eliminate \(r\) from these equations. This will typically involve substituting one equation into the other. ### Step 7: Determine the Number of Unknowns After eliminating \(r\), we will have an equation in terms of \(c\) and \(d\). The number of unknowns in this equation will determine the order of the differential equation. ### Step 8: Conclusion In this case, since \(a_1\), \(b_1\), \(r_1\), \(a_2\), \(b_2\), and \(r_2\) are known constants, the only unknowns are \(c\) and \(d\). Thus, we have two unknowns. The order of the differential equation is equal to the number of unknown parameters we need to eliminate. Since we have two unknowns, the order of the differential equation of the family of circles touching the two given circles externally is **2**. ### Final Answer The order of the differential equation of the family of circles touching two given circles externally is **2**. ---
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OBJECTIVE RD SHARMA-DIFFERENTIAL EQUATIONS-Exercise
  1. The solution of the differential equation (dy)/(dx)+(y)/(x)=x^(2), is

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  2. The solution of differential equation (1+y^(2))+(x-e^(tan^(-1)y))(dy...

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  3. Solution of x(dy)/(dx)+y=xe^(x), is

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  4. The tangent at any point (x , y) of a curve makes an angle tan^(-1)(2x...

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  5. The integrating factor of the differential equation (dy)/(dx)+y=(1+y)/...

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  6. The degree of the differential equation corresponding to the family of...

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  7. The degree of the differential equation of all curves having normal of...

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  8. The differential equation of the family of ellipses having major and m...

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  9. The degree of the differential equation satisfying the relation sqrt(1...

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  10. The differential eqaution of the family of curve y^(2)=4a(x+1), is

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  11. Find the equation of the curve in which the subnormal varies as the ...

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  12. Solution of the differential equation xdy-ydx=0 represents

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  13. The equation of the curve whose subnormal is twice the abscissa, is

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  14. The solution of the differential equation (x)/(x^(2)+y^(2))dy=((y)/(...

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  15. A curve passes through the point (0,1) and the gradient at (x,y) on it...

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  16. The equation of the curve through the point (1,0), whose slope is (y-...

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  17. The differential equation for which sin^(-1) x + sin^(-1) y = c is giv...

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  18. Solution of the differential equation (dx)/(x)+(dy)/(y)=0 is

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  19. The order of the differential equation of family of circles touching t...

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  20. The function f(x) satisfying the equation f^2 (x) + 4 f'(x) f(x) + (...

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