CBSE Class 11 Mathematics Conic Sections Assignment Set 01

Question. If a circle passes through the point (a, b) and cuts the circle x² + y² = p orthogonally, then the equation of the locus of its centre is 
(a) x² + y² – 3ax – 4by + (a² + b² - p²) = 0
(b) 2ax + 2by – (a² - b² + p²) = 0
(c) x² + y² – 2ax – 3by + (a² - b² - p² ) = 0
(d) 2ax + 2by – (a² + b² + p² ) = 0
Answer : D

Question. If the lines 3x - 4y - 7 = 0 and 2x - 3y - 5 = 0 are two diameters of a circle of area 49p square units, the equation of the circle is 
(a) x² + y² + 2x - 2y - 47 = 0
(b) x² + y² + 2x - 2y - 62 = 0
(c) x² + y² - 2x + 2y - 62 = 0
(d) x² + y² - 2x + 2y - 47 = 0
Answer : D

Question. The two circles x² + y² = ax and x² + y² = c² (c > 0) touch each other if 
(a) | a | = c
(b) a = 2c
(c) | a | = 2c
(d) 2 | a | = c
Answer : A

Question. The length of the diameter of the circle which touches the x-axis at the point (1,0) and passes through the point (2,3) is:
(a) 10/3
(b) 3/5
(c) 6/5
(d) 5/3
Answer : A

Question. The number of common tangents of the circles given by x² + y² – 8x – 2y + 1 = 0 and x² + y² + 6x + 8y = 0 is
(a) one
(b) four
(c) two
(d) three
Answer : C

Question. The circle x² + y² = 4x + 8y + 5 intersects the line 3x – 4y = m at two distinct points if 
(a) – 35 < m < 15
(b) 15 < m < 65
(c) 35 < m < 85
(d) – 85 < m < – 35
Answer : A

Question. If the pair of lines ax² + 2(a + b)xy + by² = 0 lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then 
(a) 3a² -10ab + 3b² = 0
(b) 3a² - 2ab + 3b² = 0
(c) 3a² +10ab + 3b² = 0
(d) 3a² + 2ab + 3b² = 0
Answer : D

Question. The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle having area as 154 sq.units.Then the equation of the circle is 
(a) x² + y² - 2x + 2y = 62
(b) x² + y² + 2x - 2y = 62
(c) x² + y² + 2x -2 y = 47
(d) x² + y² - 2x + 2y = .47
Answer : D

Question. The point diametrically opposite to the point P(1, 0) on the circle x² + y² + 2x + 4y – 3 = 0 is
(a) (3, – 4)
(b) (–3, 4)
(c) (–3, –4)
(d) (3, 4)
Answer : C

Question. The centre of the circle passing through (0, 0) and (1, 0) and touching the circle x² + y² = 9 is
(a) (1/2, 1/2)
(b) (1/2, - √2)
(c) (3/2, 1/2)
(d) (1/2, 3/2)
Answer : B

Question. If the chord y = mx + 1 of the circle x² + y² = 1 subtends an angle of measure 45 at the ma or segment of the circle then value of m is 
(a) 2 ± √2
(b) –2 ± √2
(c) –1 ± √2
(d) none of these
Answer : C

Question. If the circles x² + y² + 2ax + cy + a = 0 and x² + y² – 3ax + dy – 1 = 0 intersect in two distinct points P and Q then the line 5x + by – a = 0 passes through P and Q for 
(a) exactly one value of a
(b) no value of a
(c) infinitely many values of a
(d) exactly two values of a
Answer : B

Question. If a circle passes through the point (a, b) and cuts the circle x² + y² = 4orthogonally, then the locus of its centre is 
(a) 2ax - 2by - (a² + b² + 4) = 0
(b) 2ax + 2by - (a² + b² + 4) = 0
(c) 2ax - 2by + (a² + b² + 4) = 0
(d) 2ax + 2by + (a² + b² + 4) = 0
Answer : B

Question. If two vertices of an equilateral triangle are A (– a, 0) and B (a, 0), a > 0, and the third vertex C lies above x-axis then the equation of the circumcircle of ΔABC is :
(a) 3x² + 3y² – 2√3ay = 3a²
(b) 3x² + 3y² - 2ay = 3a²
(c) x² + y² - 2ay = a²
(d) x² + y² - √3ay = a²
Answer : A

Question. If P and Q are the points of intersection of the circles x² + y²+ 3x + 7y + 2p - 5 = 0 and x² + y² + 2x + 2y – p² = 0 then there is a circle passing through P, Q and (1, 1) for: 
(a) all except one value of p
(b) all except two values of p
(c) exactly one value of p
(d) all values of p
Answer : A

Question. A variable circle passes through the fixed point A( p,q) and touches x-axis . The locus of the other end of the diameter through A is 
(a) (y - q)² = 4px
(b) (x - q)² = 4 py
(c) (y - p)² = 4qx
(d) (x - p)² = 4qy
Answer : D

Question. Let C be the circle with centre (0, 0) and radius 3 units. The equation of the locus of the mid points of the chords of the circle C that subtend an angle of 2π/3 at its center is
(a) x² + y² = 3/2
(b) x² + y² = 1
(c) x² + y² = 27/4
(d) x² + y² = 9/4
Answer : D

Question. Intercept on the line y = x by the circle x² + y² - 2x = 0 is AB. Equation of the circle on AB as a diameter is 
(a) x² + y² - y = 0
(b) x² + y² - x + y = 0
(c) x² + y² + x + y = 0
(d) x² + y² - x - y = 0
Answer : D

Question. If the two circles (x -1)² + ( y - 3)² = r² and x² + y² - 8x + 2y + = 0 intersect in two distinct point, then 
(a) r > 2
(b) 2 < r < 8
(c) r < 2
(d) r = 2.
Answer : B

Question. The centres of a set of circles, each of radius 3, lie on the circle x² + y²=25. The locus of any point in the set is
(a) 4 ≤ x² + y² ≤ 64
(b) x² + y² ≤ 25
(c) x² + y² ³ 25
(d) 3 ≤ x² + y² ≤ 9
Answer : A

Question. If the lines 2x + 3y +1 = 0 and 3x - y - 4 = 0 lie along diameter of a circle of circumference 10p, then the equation of the circle is 
(a) x² + y² + 2x - 2y - 23 = 0
(b) x² + y² - 2x - 2y - 23 = 0
(c) x² + y² + 2x + 2y - 23 = 0
(d) x² + y² - 2x + 2y - 23 = 0
Answer : D

Question. The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length 3a is
(a) x² + y² = 9a²
(b) x² + y² = 16a²
(c) x² + y² = 4a²
(d) x² + y² = a²
Answer : C

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)

Question 1. Find the centre and radius of the circle \(3x^2 + 3y^2 + 6x - 4y - 1 = 0\)
Answer: Given equation of the circle is:
\(3x^2 + 3y^2 + 6x - 4y - 1 = 0\)
Dividing the entire equation by 3 to make the coefficients of \(x^2\) and \(y^2\) equal to 1:
\(x^2 + y^2 + 2x - \frac{4}{3}y - \frac{1}{3} = 0\)
Comparing this with the general equation of a circle \(x^2 + y^2 + 2gx + 2fy + c = 0\):
\(2g = 2 \implies g = 1\)
\(2f = -\frac{4}{3} \implies f = -\frac{2}{3}\)
\(c = -\frac{1}{3}\)
Centre = \((-g, -f) = \left(-1, \frac{2}{3}\right)\)
Radius \(r = \sqrt{g^2 + f^2 - c} = \sqrt{(1)^2 + \left(-\frac{2}{3}\right)^2 - \left(-\frac{1}{3}\right)}\)
\(r = \sqrt{1 + \frac{4}{9} + \frac{1}{3}} = \sqrt{\frac{9 + 4 + 3}{9}} = \sqrt{\frac{16}{9}} = \frac{4}{3}\)
Thus, the centre is \(\left(-1, \frac{2}{3}\right)\) and the radius is \(\frac{4}{3}\).
In simple words: We divided by 3, found \(g, f, c\), and used the standard centre and radius formulas.

Exam Tip: Always ensure the coefficients of \(x^2\) and \(y^2\) are 1 before finding the centre and radius.

Question 2. Does \(2x^2 + 2y^2 + 3y + 10 = 0\) represent the equation of a circle? Justify.
Answer: Given equation:
\(2x^2 + 2y^2 + 3y + 10 = 0\)
Divide by 2:
\(x^2 + y^2 + \frac{3}{2}y + 5 = 0\)
Here, \(g = 0\), \(f = \frac{3}{4}\), and \(c = 5\).
Radius \(r = \sqrt{g^2 + f^2 - c} = \sqrt{0 + \frac{9}{16} - 5} = \sqrt{-\frac{71}{16}}\)
Since the value under the square root is negative, the radius is not a real number.
Therefore, it does not represent a real circle.
In simple words: The calculated radius is imaginary, so no real circle is formed.

Exam Tip: A second-degree equation in \(x\) and \(y\) represents a real circle only if \(g^2 + f^2 - c \geq 0\).

Question 3. Find equation of circle whose end points of one of its diameter are \((-2, 3)\) and \((0, -1)\).
Answer: Using the diameter form of a circle equation with endpoints \((x_1, y_1)\) and \((x_2, y_2)\):
\((x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0\)
Substitute \((-2, 3)\) and \((0, -1)\):
\((x - (-2))(x - 0) + (y - 3)(y - (-1)) = 0\)
\((x + 2)x + (y - 3)(y + 1) = 0\)
\(x^2 + 2x + y^2 + y - 3y - 3 = 0\)
\(x^2 + y^2 + 2x - 2y - 3 = 0\)
Alternatively, in center-radius form: \((x + 1)^2 + (y - 1)^2 = 5\).
In simple words: Use the diameter endpoints formula directly to expand and get the equation.

Exam Tip: The diameter form saves time compared to finding the centre and radius first.

Question 4. Find the value(s) of p so that the equation \(x^2 + y^2 - 2px + 4y - 12 = 0\) may represent a circle of radius 5 units.
Answer: Given equation:
\(x^2 + y^2 - 2px + 4y - 12 = 0\)
Here, \(2g = -2p \implies g = -p\)
\(2f = 4 \implies f = 2\)
\(c = -12\)
Radius \(r = \sqrt{g^2 + f^2 - c} = 5\)
\(\sqrt{(-p)^2 + (2)^2 - (-12)} = 5\)
\(\sqrt{p^2 + 4 + 12} = 5\)
\(\sqrt{p^2 + 16} = 5\)
Squaring both sides:
\(p^2 + 16 = 25\)
\(p^2 = 9\)
\(p = \pm 3\)
In simple words: Substitute \(g, f, c\) into the radius formula and solve for \(p\).

Exam Tip: Remember to consider both positive and negative square roots when solving \(p^2 = 9\).

Question 5. If parabola \(y^2 = px\) passes through point \((2, -3)\), find the length of latus rectum.
Answer: Given parabola equation: \(y^2 = px\)
Since it passes through \((2, -3)\), substitute \(x = 2\) and \(y = -3\):
\((-3)^2 = p(2)\)
\(9 = 2p \implies p = \frac{9}{2}\)
The standard form is \(y^2 = 4ax\). Comparing \(y^2 = px\) with \(y^2 = 4ax\), the length of the latus rectum is \(4a = p\).
Thus, the length of the latus rectum is \(\frac{9}{2}\).
In simple words: Plug the point into the equation to find \(p\), which equals the latus rectum length.

Exam Tip: For \(y^2 = 4ax\), the coefficient of \(x\) represents the entire length of the latus rectum.

Question 6. Find the coordinates of focus, and length of latus rectum of parabola \(3y^2 = 8x\).
Answer: Given equation: \(3y^2 = 8x\)
Dividing by 3:
\(y^2 = \frac{8}{3}x\)
Comparing with the standard right-handed parabola \(y^2 = 4ax\):
\(4a = \frac{8}{3} \implies a = \frac{2}{3}\)
The focus of \(y^2 = 4ax\) is given by \((a, 0)\).
Therefore, the focus is \(\left(\frac{2}{3}, 0\right)\).
The length of the latus rectum is \(4a = \frac{8}{3}\).
In simple words: Convert to standard form, find \(a\), then write the focus and latus rectum.

Exam Tip: Always reduce the equation to standard form with coefficient 1 for \(y^2\) or \(x^2\).

Question 7. Find the eccentricity of the ellipse \(\frac{x^2}{25} + \frac{y^2}{9} = 1\)
Answer: Given ellipse equation: \(\frac{x^2}{25} + \frac{y^2}{9} = 1\)
Here, \(a^2 = 25 \implies a = 5\)
\(b^2 = 9 \implies b = 3\)
Since \(a > b\), the major axis is along the x-axis.
The relation between \(a, b, c\) is \(c = \sqrt{a^2 - b^2}\):
\(c = \sqrt{25 - 9} = \sqrt{16} = 4\)
Eccentricity \(e = \frac{c}{a} = \frac{4}{5}\)
In simple words: Find \(a\) and \(b\), calculate \(c\), and use \(e = \frac{c}{a}\).

Question 8. One end of diameter of a circle \(x^2 + y^2 - 6x + 5y - 7 = 0\) is \((7, -8)\). Find the coordinates of other end.
Answer: Given circle equation: \(x^2 + y^2 - 6x + 5y - 7 = 0\)
Find the centre \((h, k)\) of the circle:
\(2g = -6 \implies g = -3 \implies h = 3\)
\(2f = 5 \implies f = \frac{5}{2} \implies k = -\frac{5}{2}\)
Centre = \(\left(3, -\frac{5}{2}\right)\)
Let the other end of the diameter be \((x_2, y_2)\) and the given end be \((x_1, y_1) = (7, -8)\).
Since the centre is the midpoint of the diameter:
\(\frac{7 + x_2}{2} = 3 \implies 7 + x_2 = 6 \implies x_2 = -1\)
\(\frac{-8 + y_2}{2} = -\frac{5}{2} \implies -8 + y_2 = -5 \implies y_2 = 3\)
Thus, the coordinates of the other end are \((-1, 3)\).
In simple words: Find the center first, then use the midpoint formula to find the unknown endpoint.

Exam Tip: The center of a circle is always the exact midpoint of any of its diameters.

Question 9. Find the equation of the ellipse coordinates of whose foci are \((\pm 2, 0)\) and length of latus rectum is \(\frac{10}{3}\).
Answer: Since the foci are on the x-axis at \((\pm 2, 0)\), the equation of the ellipse is of the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).
Here, \(c = 2\).
Length of latus rectum = \(\frac{2b^2}{a} = \frac{10}{3} \implies \frac{b^2}{a} = \frac{5}{3} \implies b^2 = \frac{5a}{3}\)
We know that \(c^2 = a^2 - b^2\):
\((2)^2 = a^2 - \frac{5a}{3}\)
\(4 = a^2 - \frac{5a}{3}\)
Multiply by 3:
\(12 = 3a^2 - 5a\)
\(3a^2 - 5a - 12 = 0\)
Factorize the quadratic equation:
\(3a^2 - 9a + 4a - 12 = 0\)
\(3a(a - 3) + 4(a - 3) = 0\)
\((3a + 4)(a - 3) = 0\)
Since \(a > 0\), we take \(a = 3\).
Now find \(b^2\):
\(b^2 = \frac{5(3)}{3} = 5\)
Thus, the equation of the ellipse is:
\(\frac{x^2}{9} + \frac{y^2}{5} = 1\)
In simple words: Use \(c=2\), latus rectum formula, and \(c^2 = a^2 - b^2\) to form a quadratic in \(a\).

Exam Tip: Always reject negative values for length parameters like \(a\) and \(b\).

Question 10. Find the equation of ellipse with eccentricity \(\frac{3}{4}\), centre at origin, foci on y-axis and passing through point \((6, 4)\).
Answer: Since foci are on the y-axis, the equation is of the form:
\(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\)
Given eccentricity \(e = \frac{c}{a} = \frac{3}{4} \implies c = \frac{3}{4}a\)
We know \(c^2 = a^2 - b^2 \implies \left(\frac{3}{4}a\right)^2 = a^2 - b^2 \implies \frac{9}{16}a^2 = a^2 - b^2\)
\(b^2 = a^2 - \frac{9}{16}a^2 = \frac{7}{16}a^2\)
Substitute \(x = 6\) and \(y = 4\) into the ellipse equation since it passes through \((6, 4)\):
\(\frac{6^2}{b^2} + \frac{4^2}{a^2} = 1\)
\(\frac{36}{b^2} + \frac{16}{a^2} = 1\)
Substitute \(b^2 = \frac{7}{16}a^2\):
\(\frac{36}{\frac{7}{16}a^2} + \frac{16}{a^2} = 1\)
\(\frac{36 \times 16}{7a^2} + \frac{16}{a^2} = 1\)
\(\frac{576}{7a^2} + \frac{112}{7a^2} = 1\)
\(\frac{688}{7a^2} = 1 \implies 7a^2 = 688\)
Since \(b^2 = \frac{7}{16}a^2\), multiply by 16 on both sides of \(7a^2 = 688\) to get \(b^2\):
Wait, substitute directly: \(b^2 = \frac{688}{16} = 43\).
Therefore, \(a^2 = \frac{688}{7}\) and \(b^2 = 43\).
The equation is \(\frac{x^2}{43} + \frac{y^2}{\frac{688}{7}} = 1 \implies 16x^2 + 7y^2 = 688\).
In simple words: Express \(b^2\) in terms of \(a^2\), substitute the given point, and solve for \(a^2\) and \(b^2\).

Exam Tip: Be careful with fraction arithmetic when substituting parameters into the equation.

Question 11. Find the equation of hyperbola with centre at origin, transverse axis along x-axis, eccentricity \(\sqrt{5}\) and sum of lengths of whose axes is 18.
Answer: Since the transverse axis is along the x-axis, the equation is:
\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
Given eccentricity \(e = \sqrt{5}\)
We know \(c = ae \implies c = a\sqrt{5}\)
Also, \(c^2 = a^2 + b^2 \implies (a\sqrt{5})^2 = a^2 + b^2 \implies 5a^2 = a^2 + b^2 \implies b^2 = 4a^2\)
Given sum of lengths of axes is 18:
\(2a + 2b = 18 \implies a + b = 9 \implies b = 9 - a\)
Substitute \(b = 9 - a\) into \(b^2 = 4a^2\):
\((9 - a)^2 = 4a^2\)
\(81 - 18a + a^2 = 4a^2\)
\(3a^2 + 18a - 81 = 0\)
Divide by 3:
\(a^2 + 6a - 27 = 0\)
\((a + 9)(a - 3) = 0\)
Since \(a > 0\), \(a = 3\).
Then \(b = 9 - 3 = 6\), so \(b^2 = 36\) and \(a^2 = 9\).
Thus, the equation is:
\(\frac{x^2}{9} - \frac{y^2}{36} = 1 \implies 4x^2 - y^2 = 36\).
In simple words: Use eccentricity and axes length relations to set up equations for \(a\) and \(b\).

Exam Tip: The length of transverse axis is \(2a\) and conjugate axis is \(2b\).

Question 12. Two diameters of a circle are along the lines \(x - y - 9 = 0\) and \(x - 2y - 7 = 0\) and area of circle is 154 square units, find its equation.
Answer: The centre of the circle is the point of intersection of its diameters. Solve the two line equations simultaneously:
\(x - y = 9\) --- (1)
\(x - 2y = 7\) --- (2)
Subtract (2) from (1):
\(y = 2\)
Substitute \(y = 2\) into (1):
\(x - 2 = 9 \implies x = 11\)
So the centre is \((11, 2)\).
Given area of the circle is 154 square units:
\(\pi r^2 = 154\)
\(\frac{22}{7} \times r^2 = 154\)
\(r^2 = 154 \times \frac{7}{22} = 7 \times 7 = 49 \implies r = 7\)
Now, using the center-radius form \((x - h)^2 + (y - k)^2 = r^2\):
\((x - 11)^2 + (y - 2)^2 = 49\)
\(x^2 - 22x + 121 + y^2 - 4y + 4 = 49\)
\(x^2 + y^2 - 22x - 4y + 125 - 49 = 0\)
\(x^2 + y^2 - 22x - 4y + 76 = 0\)
In simple words: Find the center by intersecting diameters, find radius from area, then write the equation.

Exam Tip: Always remember \(\pi \approx \frac{22}{7}\) when calculating circle radius from area.

Question 13. Find equation(s) of circle passing through points \((1, 1)\), \((2, 2)\) and whose radius is 1 unit.
Answer: Let the equation of the circle be \(x^2 + y^2 + 2gx + 2fy + c = 0\) with radius \(r = 1\).
Since it passes through \((1, 1)\):
\(1^2 + 1^2 + 2g(1) + 2f(1) + c = 0 \implies 2g + 2f + c = -2\) --- (1)
Since it passes through \((2, 2)\):
\(2^2 + 2^2 + 2g(2) + 2f(2) + c = 0 \implies 4g + 4f + c = -8\) --- (2)
Subtract (1) from (2):
\(2g + 2f = -6 \implies g + f = -3 \implies f = -3 - g\)
Radius \(r = \sqrt{g^2 + f^2 - c} = 1 \implies g^2 + f^2 - c = 1\) --- (3)
From (1), \(c = -2 - 2g - 2f\). Substitute \(f = -3 - g\):
\(c = -2 - 2g - 2(-3 - g) = -2 - 2g + 6 + 2g = 4\)
Substitute \(c = 4\) and \(f = -3 - g\) into (3):
\(g^2 + (-3 - g)^2 - 4 = 1\)
\(g^2 + 9 + 6g + g^2 - 4 = 1\)
\(2g^2 + 6g + 5 = 1\)
\(2g^2 + 6g + 4 = 0\)
\(g^2 + 3g + 2 = 0\)
\((g + 1)(g + 2) = 0\)
Case 1: \(g = -1\)
\(f = -3 - (-1) = -2\)
Equation: \(x^2 + y^2 - 2x - 4y + 4 = 0\)
Case 2: \(g = -2\)
\(f = -3 - (-2) = -1\)
Equation: \(x^2 + y^2 - 4x - 2y + 4 = 0\)
In simple words: Substitute given points and radius condition into the general circle equation to find two possible circles.

Exam Tip: Two distinct circles can often pass through given points for a fixed radius.

Question 14. Find equation of circle concentric with circle \(4x^2 + 4y^2 - 12x - 16y - 21 = 0\) and of half its area.
Answer: Given circle: \(4x^2 + 4y^2 - 12x - 16y - 21 = 0\)
Divide by 4:
\(x^2 + y^2 - 3x - 4y - \frac{21}{4} = 0\)
Centre \((h, k) = \left(\frac{3}{2}, 2\right)\)
Radius \(r_1 = \sqrt{\left(-\frac{3}{2}\right)^2 + (-2)^2 - \left(-\frac{21}{4}\right)} = \sqrt{\frac{9}{4} + 4 + \frac{21}{4}} = \sqrt{\frac{30}{4} + 4} = \sqrt{\frac{46}{4}} = \frac{\sqrt{46}}{2}\)
Area of this circle \(A_1 = \pi r_1^2 = \pi \left(\frac{46}{4}\right) = \frac{23\pi}{2}\)
Area of the new circle \(A_2 = \frac{1}{2}A_1 = \frac{23\pi}{4}\)
Let radius of the new circle be \(r_2\):
\(\pi r_2^2 = \frac{23\pi}{4} \implies r_2^2 = \frac{23}{4}\)
Since the new circle is concentric, its centre is \(\left(\frac{3}{2}, 2\right)\).
Equation of the new circle:
\(\left(x - \frac{3}{2}\right)^2 + (y - 2)^2 = \frac{23}{4}\)
\(x^2 - 3x + \frac{9}{4} + y^2 - 4y + 4 = \frac{23}{4}\)
\(x^2 + y^2 - 3x - 4y + \frac{25}{4} = \frac{23}{4}\)
\(x^2 + y^2 - 3x - 4y + \frac{2}{4} = 0\)
\(2x^2 + 2y^2 - 6x - 8y + 1 = 0\)
In simple words: Find the center and radius of the first circle, halve the area to get the new radius, and write the equation.

Exam Tip: Concentric circles share the exact same center coordinates.

Question 15. Find the equation of a circle whose centre is at \((4, -2)\) and \(3x - 4y + 5 = 0\) is tangent to circle.
Answer: Given centre \((h, k) = (4, -2)\).
The radius of the circle is the perpendicular distance from the centre \((4, -2)\) to the tangent line \(3x - 4y + 5 = 0\).
\(r = \left| \frac{3(4) - 4(-2) + 5}{\sqrt{3^2 + (-4)^2}} \right|\)
\(r = \left| \frac{12 + 8 + 5}{\sqrt{9 + 16}} \right| = \left| \frac{25}{5} \right| = 5\)
Now, using the centre-radius form:
\((x - 4)^2 + (y - (-2))^2 = 5^2\)
\((x - 4)^2 + (y + 2)^2 = 25\)
\(x^2 - 8x + 16 + y^2 + 4y + 4 = 25\)
\(x^2 + y^2 - 8x + 4y + 20 = 25\)
\(x^2 + y^2 - 8x + 4y - 5 = 0\)
In simple words: Calculate the radius using the perpendicular distance formula from point to line, then write the circle equation.

Exam Tip: The radius is always perpendicular to the tangent line at the point of contact.

LONG ANSWER TYPE QUESTIONS (6 MARKS)

Question 16. Show that the four points \((7, 5), (6, -2), (-1, -1)\) and \((0, 6)\) are concyclic. [Concyclic points : Four or more points which lie on a circle].
Answer: Let the general equation of a circle be:
\(x^2 + y^2 + 2gx + 2fy + c = 0\)
Since the circle passes through \((7, 5)\):
\(7^2 + 5^2 + 2g(7) + 2f(5) + c = 0\)
\(49 + 25 + 14g + 10f + c = 0 \implies 14g + 10f + c = -74\) --- (1)
Passing through \((6, -2)\):
\(6^2 + (-2)^2 + 2g(6) + 2f(-2) + c = 0\)
\(36 + 4 + 12g - 4f + c = 0 \implies 12g - 4f + c = -40\) --- (2)
Passing through \((-1, -1)\):
\((-1)^2 + (-1)^2 + 2g(-1) + 2f(-1) + c = 0\)
\(1 + 1 - 2g - 2f + c = 0 \implies -2g - 2f + c = -2\) --- (3)
Subtract (2) from (1):
\(2g + 14f = -34 \implies g + 7f = -17\) --- (4)
Subtract (3) from (2):
\(14g - 2f = -38 \implies 7g - f = -19\) --- (5)
Multiply (5) by 7:
\(49g - 7f = -133\) --- (6)
Add (4) and (6):
\(50g = -150 \implies g = -3\)
Substitute \(g = -3\) into (4):
\(-3 + 7f = -17 \implies 7f = -14 \implies f = -2\)
Substitute \(g = -3\) and \(f = -2\) into (3):
\(-2(-3) - 2(-2) + c = -2\)
\(6 + 4 + c = -2 \implies 10 + c = -2 \implies c = -12\)
Thus, the equation of the circle passing through the first three points is:
\(x^2 + y^2 - 6x - 4y - 12 = 0\)
Now, check if the fourth point \((0, 6)\) satisfies this equation:
Substitute \(x = 0\) and \(y = 6\):
\(0^2 + 6^2 - 6(0) - 4(6) - 12 = 0 + 36 - 0 - 24 - 12 = 36 - 36 = 0\)
Since the fourth point satisfies the circle equation, all four points are concyclic.
In simple words: Find the circle equation using three points, then check if the fourth point satisfies it.

Exam Tip: Concyclic proof always requires finding the circle from three points and verifying the fourth point.