The equation of a circle which passes through the points of intersection of the circles \[ 2x^2 + 2y^2 - 2x + 6y - 3 = 0, \quad x^2 + y^2 + 4x + 2y + 1 = 0 \] and whose centre lies on the common chord of these circles is:
The equation of a circle which passes through the points of intersection of the circles \[ 2x^2 + 2y^2 - 2x + 6y - 3 = 0, \quad x^2 + y^2 + 4x + 2y + 1 = 0 \] and whose centre lies on the common chord of these circles is:
Show Hint
- \( 2x^2 + 2y^2 - 3x + 4y - 2 = 0 \)
- \( x^2 + y^2 + 2x + 5y - 2 = 0 \)
- \( 3x^2 + 3y^2 - 2x + 4y - 3 = 0 \)
\( 4x^2 + 4y^2 + 6x + 10y - 1 = 0 \)
The Correct Option is D
Solution and Explanation
Step 1: Finding the Equation of the Common Chord
The given circles are: \[ 2x^2 + 2y^2 - 2x + 6y - 3 = 0, \] \[ x^2 + y^2 + 4x + 2y + 1 = 0. \] Subtracting the second equation from the first: \[ (2x^2 + 2y^2 - 2x + 6y - 3) - (x^2 + y^2 + 4x + 2y + 1) = 0. \] \[ x^2 + y^2 - 6x + 4y - 4 = 0. \]
Step 2: Finding the Required Circle
A circle passing through the intersection of these two given circles is of the form: \[ S_1 + \lambda S_2 = 0. \] Substituting and simplifying using the condition that the center lies on the common chord, we derive: \[ 4x^2 + 4y^2 + 6x + 10y - 1 = 0. \]
Step 3: Conclusion
Thus, the final answer is: \[ \boxed{4x^2 + 4y^2 + 6x + 10y - 1 = 0}. \]
Top Questions on Geometry
- The equation of chord AB of ellipse \(2x^2 + y^2 = 1\) is \(x - y + 1 = 0\). If O is the origin, then \(\angle AOB =\)
In the given figure, the numbers associated with the rectangle, triangle, and ellipse are 1, 2, and 3, respectively. Which one among the given options is the most appropriate combination of \( P \), \( Q \), and \( R \)?
- Let the sum of the focal distances of the point $ P(4, 3) $ on the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be $ 8\sqrt{\frac{5}{3}} $. If for $ H $, the length of the latus rectum is $ \ell $ and the product of the focal distances of the point $ P $ is $ m $, then $ 9\ell^2 + 6m $ is equal to:
- Let for two distinct values of $ p $, the lines $ y = x + p $ touch the ellipse $ E: \frac{x^2}{4} + \frac{y^2}{9} = 1 $ at the points $ A $ and $ B $. Let the line $ y = x $ intersect $ E $ at the points $ C $ and $ D $. Then the area of the quadrilateral $ ABCD $ is equal to:
The center of a circle $ C $ is at the center of the ellipse $ E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, where $ a>b $. Let $ C $ pass through the foci $ F_1 $ and $ F_2 $ of $ E $ such that the circle $ C $ and the ellipse $ E $ intersect at four points. Let $ P $ be one of these four points. If the area of the triangle $ PF_1F_2 $ is 30 and the length of the major axis of $ E $ is 17, then the distance between the foci of $ E $ is:
Questions Asked in TS EAMCET exam
- $7\hat{i}-4\hat{j}+7\hat{k}, \hat{i}-6\hat{j}+10\hat{k}, -\hat{i}-3\hat{j}+4\hat{k}, 5\hat{i}-\hat{j}+\hat{k}$ are the position vectors of the points A, B, C, D respectively. If $p\hat{i} + q\hat{j} + r\hat{k}$ is the position vector of the point of intersection of the diagonals of the quadrilateral ABCD, then $p+q+r=$
- TS EAMCET - 2025
- Vector Algebra
- A particle is executing simple harmonic motion with an amplitude of 10 cm. If the kinetic energy of the particle at a distance of 6 cm from the mean position is 100 J, then the kinetic energy of the particle at a distance of 2 cm from the mean position is
- TS EAMCET - 2025
- Waves and Oscillations
- If the sum of two vectors is a unit vector, then the magnitude of their difference is:
- Which of the following is an example of antifertility drug?
- TS EAMCET - 2025
- Chemistry in Everyday Life
In a messenger RNA molecule, untranslated regions (UTRs) are present at:
I. 5' end before start codon
II. 3' end after stop codon
III. 3' end before stop codon
IV. 5' end after start codon- TS EAMCET - 2025
- Plant Ecology