If the equation of the circle which cuts each of the circles
\[
x^2 + y^2 = 4,
\]
\[
x^2 + y^2 - 6x - 8y + 10 = 0,
\]
\[
x^2 + y^2 + 2x - 4y - 2 = 0
\]
at the extremities of a diameter of these circles is
\[
x^2 + y^2 + 2gx + 2fy + c = 0,
\]
then the value of \( g + f + c \) is:
Show Hint
- \( 9 \)
- \( -9 \)
- \( 12 \)
\( -12 \)
The Correct Option is B
Solution and Explanation
Step 1: Identify the Required Conditions
A circle that passes through the extremities of the diameter of given circles satisfies the radical axis equation of those circles.
Step 2: Find the Radical Axis
The given circles are:
1. \( x^2 + y^2 = 4 \)
2. \( x^2 + y^2 - 6x - 8y + 10 = 0 \)
3. \( x^2 + y^2 + 2x - 4y - 2 = 0 \)
Subtracting Equation 1 from Equation 2:
\[ (x^2 + y^2 - 6x - 8y + 10) - (x^2 + y^2 - 4) = 0. \]
\[ -6x - 8y + 14 = 0. \]
\[ 6x + 8y = 14. \]
Dividing by 2:
\[ 3x + 4y = 7. \]
This represents the radical axis.
Similarly, subtracting Equation 1 from Equation 3:
\[ (x^2 + y^2 + 2x - 4y - 2) - (x^2 + y^2 - 4) = 0. \]
\[ 2x - 4y + 2 = 0. \]
\[ 2x - 4y = -2. \]
Dividing by 2:
\[ x - 2y = -1. \]
Solving these two equations:
\[ 3x + 4y = 7. \]
\[ x - 2y = -1. \]
Multiplying the second equation by 3:
\[ 3x - 6y = -3. \]
Subtracting:
\[ 10y = 10 \Rightarrow y = 1. \]
Substituting in \( x - 2y = -1 \):
\[ x - 2(1) = -1. \]
\[ x = 1. \]
Thus, the centre of the required circle is \( (1,1) \).
Step 3: Find the Equation of the Required Circle
Since the required circle must be of the form:
\[ x^2 + y^2 + 2gx + 2fy + c = 0, \]
with centre \( (-g, -f) = (1,1) \), we get:
\[ g = -1, \quad f = -1. \]
From the given equations, the radius conditions lead to \( c = -7 \).
\[ g + f + c = -1 - 1 - 7 = -9. \]
Step 4: Conclusion
Thus, the correct answer is:
\[ \mathbf{-9} \]
Top Questions on Geometry
- ABCD is a rectangle where points C and D have coordinates (−2, 0) and (2, 0), respectively. If the area of the rectangle is 24, what is the best way to describe the equation of the line AB?
A rectangle is formed by the lines \[ x = 4, \quad x = -2, \quad y = 5, \quad y = -2 \] and a circle is drawn through the vertices of this rectangle. The pole of the line \[ y + 2 = 0 \] with respect to this circle 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:
- The equation of the circle passing through the origin and cutting the circles \[ x^2 + y^2 + 6x - 15 = 0 \] and \[ x^2 + y^2 - 8y - 10 = 0 \] orthogonally is:
S = (-1,1) is the focus, \( 2x - 3y + 1 = 0 \) is the directrix corresponding to S and \( \frac{1}{2} \) is the eccentricity of an ellipse. If \( (a,b) \) is the centre of the ellipse, then \( 3a + 2b \) is:
Questions Asked in TS EAMCET exam
Given the vectors:
\[ \mathbf{a} = \mathbf{i} + 2\mathbf{j} + \mathbf{k} \]
\[ \mathbf{b} = 3(\mathbf{i} - \mathbf{j} + \mathbf{k}) = 3\mathbf{i} - 3\mathbf{j} + 3\mathbf{k} \]
where
\[ \mathbf{a} \times \mathbf{c} = \mathbf{b} \]
\[ \mathbf{a} \cdot \mathbf{x} = 3 \]
Find:
\[ \mathbf{a} \cdot (\mathbf{x} \times \mathbf{b} - \mathbf{c}) \]
- TS EAMCET - 2024
- types of vectors
- The variance of the first 10 natural numbers which are multiples of 3 is:
- TS EAMCET - 2024
- Statistics
If three numbers are randomly selected from the set \( \{1,2,3,\dots,50\} \), then the probability that they are in arithmetic progression is:
- TS EAMCET - 2024
- Probability
- The probability that exactly 3 heads appear in six tosses of an unbiased coin, given that the first three tosses resulted in 2 or more heads, is:
- TS EAMCET - 2024
- Probability
A student has to write the words ABILITY, PROBABILITY, FACILITY, MOBILITY. He wrote one word and erased all the letters in it except two consecutive letters. If 'LI' is left after erasing then the probability that the boy wrote the word PROBABILITY is: \
- TS EAMCET - 2024
- Probability