Q1. centre (0,2) and radius 2

Solution

Equation of a circle with centre at (0, 2) and radius 2

The standard equation of a circle with centre \((h, k)\) and radius \(r\) is \[ \begin{aligned} (x-h)^2 + (y-k)^2 = r^2 \end{aligned} \] This form directly represents the geometric definition of a circle as the locus of a point at a fixed distance from a fixed centre.

Here, the given centre is \((0, 2)\), so \(h = 0\) and \(k = 2\), and the radius is \(r = 2\). Substituting these values into the standard equation, we get \[ \begin{aligned} (x-0)^2 + (y-2)^2 = 2^2 \\ x^2 + (y-2)^2 = 4 \end{aligned} \]

Expanding the square and simplifying further, \[ \begin{aligned} x^2 + y^2 - 4y + 4 = 4 \\ x^2 + y^2 - 4y = 0 \end{aligned} \] This is the required equation of the circle.

Q2. centre (–2,3) and radius 4

Solution

Equation of a circle with centre at (−2, 3) and radius 4

The general equation of a circle with centre \((h, k)\) and radius \(r\) is \[ \begin{aligned} (x-h)^2 + (y-k)^2 = r^2 \end{aligned} \] This equation follows directly from the definition of a circle as the set of all points whose distance from a fixed point is constant.

In the given problem, the centre is \((-2, 3)\), so \(h = -2\) and \(k = 3\), and the radius is \(r = 4\). Substituting these values into the standard equation, we obtain \[ \begin{aligned} (x-(-2))^2 + (y-3)^2 = 4^2 \\ (x+2)^2 + (y-3)^2 = 16 \end{aligned} \]

Expanding the squares and simplifying, \[ \begin{aligned} x^2 + 4x + 4 + y^2 - 6y + 9 = 16 \\ x^2 + y^2 + 4x - 6y - 3 = 0 \end{aligned} \] This is the required equation of the circle in general form.

Q3. centre \((\dfrac{1}{2},\dfrac{1}{4})\) and radius \(\dfrac{1}{12}\)

Solution

Equation of a circle with centre \(\left(\dfrac{1}{2}, \dfrac{1}{4}\right)\) and radius \(\dfrac{1}{12}\)

The standard equation of a circle with centre \((h, k)\) and radius \(r\) is \[ \begin{aligned} (x-h)^2 + (y-k)^2 = r^2 \end{aligned} \] This equation is obtained from the distance formula by fixing the distance of any point on the circle from its centre.

Here, the given centre is \(\left(\dfrac{1}{2}, \dfrac{1}{4}\right)\), so \(h = \dfrac{1}{2}\) and \(k = \dfrac{1}{4}\), and the radius is \(r = \dfrac{1}{12}\). Substituting these values in the standard equation, we get \[ \begin{aligned} \left(x-\dfrac{1}{2}\right)^2 + \left(y-\dfrac{1}{4}\right)^2 = \left(\dfrac{1}{12}\right)^2 \end{aligned} \]

Expanding and simplifying, \[ \begin{aligned} x^2 - x + \dfrac{1}{4} + y^2 - \dfrac{y}{2} + \dfrac{1}{16} = \dfrac{1}{144} \\ x^2 + y^2 - x - \dfrac{y}{2} = \dfrac{1}{144} - \dfrac{1}{16} - \dfrac{1}{4} \\ x^2 + y^2 - x - \dfrac{y}{2} + \dfrac{11}{36} = 0 \end{aligned} \]

To remove fractions, multiplying the entire equation by \(36\), \[ \begin{aligned} 36x^2 + 36y^2 - 36x - 18y + 11 = 0 \end{aligned} \] This is the required equation of the circle in general form.

Q4. centre (1,1) and radius \(\sqrt{2}\)

Solution

Equation of a circle with centre at (1, 1) and radius \(\sqrt{2}\)

The standard equation of a circle with centre \((h, k)\) and radius \(r\) is \[ \begin{aligned} (x-h)^2 + (y-k)^2 = r^2 \end{aligned} \] This equation represents all points whose distance from the given centre remains constant.

In this case, the centre is \((1, 1)\), so \(h = 1\) and \(k = 1\), and the radius is \(r = \sqrt{2}\). Substituting these values into the standard equation, we obtain \[ \begin{aligned} (x-1)^2 + (y-1)^2 = (\sqrt{2})^2 \end{aligned} \]

Expanding both squares and simplifying, \[ \begin{aligned} x^2 - 2x + 1 + y^2 - 2y + 1 &= 2 \\ x^2 + y^2 - 2(x + y) &= 0 \end{aligned} \] This is the required equation of the circle in general form.

Q5. centre \((–a, –b)\) and radius \(\sqrt{a_2-b^2}\).

Solution

Equation of a circle with centre \(( -a, -b )\) and radius \(\sqrt{a^{2}-b^{2}}\)

The standard equation of a circle with centre \((h, k)\) and radius \(r\) is \[ \begin{aligned} (x-h)^2 + (y-k)^2 = r^2 \end{aligned} \] This equation follows from the definition of a circle as the locus of a point whose distance from a fixed centre remains constant.

Here, the centre is given as \((-a, -b)\), so \(h = -a\) and \(k = -b\), and the radius is \(r = \sqrt{a^{2}-b^{2}}\). Substituting these values in the standard equation, we get \[ \begin{aligned} (x-(-a))^2 + (y-(-b))^2 & (\sqrt{a^{2}-b^{2}})^2 \\ (x+a)^2 + (y+b)^2 &= a^{2}-b^{2} \end{aligned} \]

Expanding and simplifying, \[ \begin{aligned} x^2 + 2ax + a^2 + y^2 + 2by + b^2 &= a^{2}-b^{2} \\ x^2 + y^2 + 2ax + 2by + a^2 + b^2 - a^2 + b^2 &= 0 \\ x^2 + y^2 + 2ax + 2by + 2b^2 &= 0 \end{aligned} \] This is the required equation of the circle in general form.

Q6. \((x + 5)^2 + (y – 3)^2 = 36\)

Solution

Centre and radius of the given circle

The given equation of the circle is \[ \begin{aligned} (x+5)^2 + (y-3)^2 = 36 \end{aligned} \] The standard form of the equation of a circle with centre \((h, k)\) and radius \(r\) is \[ \begin{aligned} (x-h)^2 + (y-k)^2 = r^2 \end{aligned} \] Comparing the given equation with the standard form helps in identifying the centre and the radius directly.

From \((x+5)^2 = (x-(-5))^2\), we get \(-h = 5\), which gives \(h = -5\). Similarly, from \((y-3)^2\), we have \(k = 3\). Also, \(r^2 = 36\), so the radius is obtained by taking the positive square root of 36. \[ \begin{aligned} r^2 = 36 \\ r = \sqrt{36} \\ r = 6 \end{aligned} \]

Hence, the centre of the circle is \((-5, 3)\) and the radius is \(6\).

Q7. \(x^2 + y^2 – 4x – 8y – 45 = 0\)

Solution

Centre and radius of the given circle

The given equation of the circle is \[ \begin{aligned} x^2 + y^2 - 4x - 8y - 45 = 0 \end{aligned} \] To find the centre and radius, the equation is converted into the standard form by completing the squares of the \(x\) and \(y\) terms.

Grouping the \(x\) terms and the \(y\) terms, \[ \begin{aligned} x^2 - 4x + y^2 - 8y - 45 = 0 \\ x^2 - 4x + 4 - 4 + y^2 - 8y + 16 - 16 - 45 = 0 \end{aligned} \] The added terms allow the formation of perfect squares.

Rewriting the equation, \[ \begin{aligned} (x-2)^2 + (y-4)^2 - 65 = 0 \\ (x-2)^2 + (y-4)^2 = 65 \end{aligned} \] This equation is now in the standard form of a circle.

Comparing with \[ \begin{aligned} (x-h)^2 + (y-k)^2 = r^2 \end{aligned} \] we obtain \(h = 2\), \(k = 4\), and \(r^2 = 65\), so the radius is \(\sqrt{65}\).

Hence, the centre of the circle is \((2, 4)\) and the radius is \(\sqrt{65}\).

Q8. \(x^2 + y^2 – 8x + 10y – 12 = 0\)

Solution

Centre and radius of the given circle

The given equation of the circle is \[ \begin{aligned} x^2 + y^2 - 8x + 10y - 12 = 0 \end{aligned} \] To determine the centre and radius, the equation is rewritten in standard form by completing the squares for the \(x\) and \(y\) terms.

Grouping the \(x\) terms and the \(y\) terms, \[ \begin{aligned} x^2 - 8x + y^2 + 10y - 12 = 0 \\ x^2 - 2\cdot4x + y^2 + 2\cdot5y - 12 = 0 \end{aligned} \] Adding and subtracting the required squares, \[ \begin{aligned} x^2 - 2\cdot4x + 16 - 16 + y^2 + 2\cdot5y + 25 - 25 - 12 = 0 \end{aligned} \]

Rewriting in terms of perfect squares, \[ \begin{aligned} (x-4)^2 + (y+5)^2 - 53 = 0 \\ (x-4)^2 + (y+5)^2 = 53 \end{aligned} \] This equation is now in the standard form of a circle.

Comparing with \[ \begin{aligned} (x-h)^2 + (y-k)^2 = r^2 \end{aligned} \] we obtain \(h = 4\), \(k = -5\), and \(r^2 = 53\), so the radius is \(\sqrt{53}\).

Hence, the centre of the circle is \((4, -5)\) and the radius is \(\sqrt{53}\).

Q9. \(2x^2 + 2y^2 – x = 0\)

Solution

Centre and radius of the given circle

The given equation of the circle is \[ \begin{aligned} 2x^2 + 2y^2 - x = 0 \end{aligned} \] First, divide the entire equation by \(2\) to simplify the coefficients, which gives \[ \begin{aligned} x^2 + y^2 - \dfrac{x}{2} = 0 \end{aligned} \]

To convert this equation into the standard form, complete the square for the \(x\)-term while keeping the \(y\)-term unchanged, \[ \begin{aligned} x^2 - \dfrac{x}{2} + y^2 = 0 \\ x^2 - \dfrac{2x}{4} + \dfrac{1}{16} - \dfrac{1}{16} + y^2 = 0 \end{aligned} \] The term \(\dfrac{1}{16}\) is added and subtracted to form a perfect square.

Rewriting the equation, \[ \begin{aligned} \left(x - \dfrac{1}{4}\right)^2 + y^2 = \dfrac{1}{16} \end{aligned} \] This is now in the standard form of a circle.

Comparing with \[ \begin{aligned} (x-h)^2 + (y-k)^2 = r^2 \end{aligned} \] we obtain \(h = \dfrac{1}{4}\), \(k = 0\), and \(r^2 = \dfrac{1}{16}\), so the radius is \[ \begin{aligned} r = \sqrt{\dfrac{1}{16}} = \dfrac{1}{4} \end{aligned} \]

Hence, the centre of the circle is \(\left(\dfrac{1}{4}, 0\right)\) and the radius is \(\dfrac{1}{4}\).

Q10. Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x + y = 16.

Solution

Equation of the required circle

Let the centre of the circle be \((h, k)\) and the radius be \(r\). The standard equation of a circle is \[ \begin{aligned} (x-h)^2 + (y-k)^2 = r^2 \end{aligned} \] Since the circle passes through the points \((4, 1)\) and \((6, 5)\), both points satisfy this equation.

Substituting \((4, 1)\) and \((6, 5)\) respectively, \[ \begin{aligned} (4-h)^2 + (1-k)^2 = r^2 \\ (6-h)^2 + (5-k)^2 = r^2 \end{aligned} \] Equating these two expressions eliminates \(r^2\), \[ \begin{aligned} (4-h)^2 + (1-k)^2 = (6-h)^2 + (5-k)^2 \end{aligned} \]

Expanding and simplifying, \[ \begin{aligned} 16 + h^2 - 8h + 1 + k^2 - 2k = 36 - 12h + h^2 + 25 - 10k + k^2 \\ -8h - 2k + 12h + 10k + 16 + 1 - 36 - 25 = 0 \\ 4h + 8k - 44 = 0 \\ h + 2k - 11 = 0 \end{aligned} \]

The centre also lies on the line \(4x + y = 16\), so \((h, k)\) satisfies \[ \begin{aligned} 4h + k = 16 \end{aligned} \] Solving the two equations, \[ \begin{aligned} k = 16 - 4h \\ h + 2(16 - 4h) - 11 = 0 \\ h + 32 - 8h - 11 = 0 \\ -7h + 21 = 0 \\ h = 3 \\ k = 16 - 4 \times 3 = 4 \end{aligned} \]

The radius is the distance between the centre \((3, 4)\) and any point on the circle, say \((4, 1)\), \[ \begin{aligned} r^2 = (4-3)^2 + (1-4)^2 \\ = 1 + 9 \\ = 10 \end{aligned} \]

Substituting \(h = 3\), \(k = 4\), and \(r^2 = 10\) into the standard equation, \[ \begin{aligned} (x-3)^2 + (y-4)^2 = 10 \\ x^2 + y^2 - 6x - 8y + 9 + 16 - 10 = 0 \\ x^2 + y^2 - 6x - 8y + 15 = 0 \end{aligned} \] This is the required equation of the circle.

Q11. Find the equation of the circle passing through the points (2,3) and (–1,1) and whose centre is on the line x – 3y – 11 = 0.

Solution

Equation of the required circle

Let the centre of the circle be \((h, k)\) and the radius be \(r\). The standard equation of a circle is \[ \begin{aligned} (x-h)^2 + (y-k)^2 = r^2 \end{aligned} \] Since the circle passes through the points \((2, 3)\) and \((-1, 1)\), both points satisfy this equation.

Substituting \((2, 3)\) and \((-1, 1)\), \[ \begin{aligned} (2-h)^2 + (3-k)^2 = r^2 \\ (-1-h)^2 + (1-k)^2 = r^2 \\ (h+1)^2 + (k-1)^2 = r^2 \end{aligned} \] Equating the two expressions for \(r^2\), \[ \begin{aligned} (h+1)^2 + (k-1)^2 = (2-h)^2 + (3-k)^2 \end{aligned} \]

Expanding and simplifying, \[ \begin{aligned} h^2 + 2h + 1 + k^2 - 2k + 1 = 4 + h^2 - 4h + 9 + k^2 - 6k \\ 2h + 4h - 2k + 6k + 2 - 13 = 0 \\ 6h + 4k - 11 = 0 \\ 6h + 4k = 11 \end{aligned} \]

The centre also lies on the line \(x - 3y - 11 = 0\), so \((h, k)\) satisfies \[ \begin{aligned} h - 3k - 11 = 0 \\ h = 3k + 11 \end{aligned} \] Substituting this value of \(h\) in \(6h + 4k = 11\), \[ \begin{aligned} 6(3k + 11) + 4k = 11 \\ 18k + 66 + 4k = 11 \\ 22k = -55 \\ k = -\dfrac{5}{2} \end{aligned} \]

Substituting \(k = -\dfrac{5}{2}\) in \(h = 3k + 11\), \[ \begin{aligned} h = 3\left(-\dfrac{5}{2}\right) + 11 \\ h = \dfrac{-15 + 22}{2} \\ h = \dfrac{7}{2} \end{aligned} \]

The radius is the distance between the centre \(\left(\dfrac{7}{2}, -\dfrac{5}{2}\right)\) and the point \((2, 3)\), \[ \begin{aligned} r^2 = \left(\dfrac{7}{2} - 2\right)^2 + \left(-\dfrac{5}{2} - 3\right)^2 \\ = \left(\dfrac{3}{2}\right)^2 + \left(\dfrac{-11}{2}\right)^2 \\ = \dfrac{9 + 121}{4} \\ = \dfrac{130}{4} \\ = \dfrac{65}{2} \end{aligned} \]

Substituting \(h = \dfrac{7}{2}\), \(k = -\dfrac{5}{2}\), and \(r^2 = \dfrac{65}{2}\) into the standard equation, \[ \begin{aligned} \left(x - \dfrac{7}{2}\right)^2 + \left(y + \dfrac{5}{2}\right)^2 = \dfrac{65}{2} \\ x^2 + y^2 - 7x + 5y = \dfrac{65}{2} - \left(\dfrac{49}{4} + \dfrac{25}{4}\right) \\ x^2 + y^2 - 7x + 5y = \dfrac{130 - 74}{4} \\ x^2 + y^2 - 7x + 5y - 14 = 0 \end{aligned} \] This is the required equation of the circle.

Q12. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3).

Solution

Equation of the required circle

Let the centre of the circle be \((h, 0)\) since it lies on the \(x\)-axis, and let the radius be \(5\). The standard equation of a circle with centre \((h, k)\) and radius \(r\) is \[ \begin{aligned} (x-h)^2 + (y-k)^2 = r^2 \end{aligned} \]

Here, \(k = 0\) and \(r = 5\). Since the circle passes through the point \((2, 3)\), this point must satisfy the equation of the circle. Substituting \((2, 3)\), \[ \begin{aligned} (2-h)^2 + (3-0)^2 = 25 \\ (2-h)^2 + 9 = 25 \\ (2-h)^2 = 16 \end{aligned} \]

Taking square roots, \[ \begin{aligned} 2-h = \pm 4 \end{aligned} \] which gives \[ \begin{aligned} h = -2 \quad \text{or} \quad h = 6 \end{aligned} \] Thus, there are two possible centres on the \(x\)-axis.

For \(h = -2\), the equation of the circle is \[ \begin{aligned} (x+2)^2 + y^2 = 25 \end{aligned} \]

For \(h = 6\), the equation of the circle is \[ \begin{aligned} (x-6)^2 + y^2 = 25 \end{aligned} \]

Hence, the required equations of the circle are \((x+2)^2 + y^2 = 25\) and \((x-6)^2 + y^2 = 25\).

Q13. Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.

Solution

Equation of the required circle

Let the circle intersect the positive \(x\)-axis at \(A(a, 0)\) and the positive \(y\)-axis at \(B(0, b)\). Since the circle passes through the origin \(O(0, 0)\), the three points \(O\), \(A\), and \(B\) lie on the circle.

The general equation of a circle is \[ \begin{aligned} x^2 + y^2 + 2gx + 2fy + c = 0 \end{aligned} \] Substituting the point \((0, 0)\), \[ \begin{aligned} c = 0 \end{aligned} \] So the equation reduces to \[ \begin{aligned} x^2 + y^2 + 2gx + 2fy = 0 \end{aligned} \]

Since the circle passes through \(A(a, 0)\), \[ \begin{aligned} a^2 + 2ga = 0 \\ 2g = -a \end{aligned} \] Similarly, since the circle passes through \(B(0, b)\), \[ \begin{aligned} b^2 + 2fb = 0 \\ 2f = -b \end{aligned} \]

Substituting the values of \(2g\) and \(2f\) in the equation of the circle, \[ \begin{aligned} x^2 + y^2 - ax - by = 0 \end{aligned} \]

Hence, the required equation of the circle passing through the origin and making intercepts \(a\) and \(b\) on the coordinate axes is \[ \begin{aligned} x^2 + y^2 - ax - by = 0 \end{aligned} \]

Q14. Find the equation of a circle with centre (2,2) and passes through the point (4,5).

Solution

Equation of the required circle

The centre of the circle is given as \((2, 2)\). The standard equation of a circle with centre \((h, k)\) and radius \(r\) is \[ \begin{aligned} (x-h)^2 + (y-k)^2 = r^2 \end{aligned} \]

Since the circle passes through the point \((4, 5)\), the radius is equal to the distance between the centre \((2, 2)\) and this point. Hence, \[ \begin{aligned} r^2 &= (4-2)^2 + (5-2)^2 \\ &= 2^2 + 3^2 \\ &= 4 + 9 \\ &= 13 \end{aligned} \]

Substituting \(h = 2\), \(k = 2\), and \(r^2 = 13\) in the standard equation of the circle, \[ \begin{aligned} (x-2)^2 + (y-2)^2 = 13 \end{aligned} \]

Thus, the required equation of the circle is \((x-2)^2 + (y-2)^2 = 13\).

Q15. Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?

Solution

Verification of the position of the given point

The given equation of the circle is \[ \begin{aligned} x^2 + y^2 = 25 \end{aligned} \] From this equation, the centre of the circle is \((0, 0)\) and the radius is \(5\).

To determine whether the point \((-2.5, 3.5)\) lies inside, outside, or on the circle, we substitute its coordinates into the left-hand side of the equation \(x^2 + y^2\), \[ \begin{aligned} (-2.5)^2 + (3.5)^2 &= 6.25 + 12.25 \\ &= 18.5 \end{aligned} \]

Now compare this value with \(25\), \[ \begin{aligned} 18.5 < 25 \end{aligned} \] Since the value obtained is less than the square of the radius, the distance of the point from the centre is less than the radius of the circle.

Hence, the point \((-2.5, 3.5)\) lies inside the given circle.