RBSE Solutions Class 11 Maths Chapter 12 Conic Section Exercise 12.1

Question 1. Find the equation of the circle whose:
(i) Centre (- 2, 3) and radius is 4
(ii) Centre (a, b) and radius is a - b
Answer:
(i) Given the information:
The center of the circle is at \( (h, k) = (-2, 3) \).
The radius is \( r = 4 \) units.
We use the standard formula for a circle's equation: \( (x - h)^2 + (y - k)^2 = r^2 \).
Substitute the values of h, k, and r into the formula:
\( [x - (-2)]^2 + (y - 3)^2 = 4^2 \)
\( \implies (x + 2)^2 + (y - 3)^2 = 16 \)
Now, expand the squared terms:
\( \implies x^2 + 4x + 4 + y^2 - 6y + 9 = 16 \)
Rearrange the terms to get the general equation of the circle:
\( \implies x^2 + y^2 + 4x - 6y + 13 = 16 \)
\( \implies x^2 + y^2 + 4x - 6y + 13 - 16 = 0 \)
\( \implies x^2 + y^2 + 4x - 6y - 3 = 0 \)
(ii) Given the information:
The center of the circle is at \( (h, k) = (a, b) \).
The radius is \( r = (a - b) \) units.
We use the standard formula for a circle's equation: \( (x - h)^2 + (y - k)^2 = r^2 \).
Substitute the values of h, k, and r into the formula:
\( (x - a)^2 + (y - b)^2 = (a - b)^2 \)
Now, expand the squared terms on both sides:
\( \implies x^2 - 2ax + a^2 + y^2 - 2by + b^2 = a^2 - 2ab + b^2 \)
Move all terms to one side to simplify the equation:
\( \implies x^2 + y^2 - 2ax - 2by + a^2 + b^2 - a^2 + 2ab - b^2 = 0 \)
Cancel out the `\( a^2 \)` and `\( b^2 \)` terms:
\( \implies x^2 + y^2 - 2ax - 2by + 2ab = 0 \)
In simple words: To find the equation of a circle, use its center coordinates and radius in the standard formula. Then expand and rearrange the terms to get the final equation.

🎯 Exam Tip: Remember the standard form of a circle's equation `\( (x - h)^2 + (y - k)^2 = r^2 \)` and how to expand binomial squares correctly. Pay close attention to signs, especially when `\( h \)` or `\( k \)` are negative.

Question 2. Find the coordinates of centre and radius of the following circles :
(i) \( x(x + y - 6) = y (x - y + 8) \)
(ii) \( \sqrt {1+k^{2}} (x^2 + y^2) = 2ax + 2aky \)
(iii) \( 4(x^2 + y^2) = 1 \)
Answer:
(i) Given the equation of the circle:
\( x(x + y - 6) = y(x - y + 8) \)
First, expand both sides of the equation:
\( \implies x^2 + xy - 6x = xy - y^2 + 8y \)
Rearrange all terms to one side to get the general form of the circle's equation:
\( \implies x^2 + y^2 + xy - xy - 6x - 8y = 0 \)
Simplify the equation:
\( \implies x^2 + y^2 - 6x - 8y = 0 \)
We compare this equation with the general form of a circle's equation, which is \( x^2 + y^2 + 2gx + 2fy + c = 0 \).
By comparing the coefficients of x, y, and the constant term, we find:
For x-term: \( 2g = -6 \implies g = -3 \)
For y-term: \( 2f = -8 \implies f = -4 \)
For the constant term: \( c = 0 \)
The coordinates of the centre \( (h, k) \) are \( (-g, -f) \).
So, the centre is \( (-(-3), -(-4)) = (3, 4) \).
The radius \( r \) is found using the formula \( r = \sqrt{g^2 + f^2 - c} \).
\( \implies r = \sqrt{(-3)^2 + (-4)^2 - 0} \)
\( \implies r = \sqrt{9 + 16 - 0} \)
\( \implies r = \sqrt{25} \)
\( \implies r = 5 \)
So, the centre is \( (3, 4) \) and the radius is \( 5 \) units.
(ii) Given the equation of the circle:
\( \sqrt{1+k^2} (x^2 + y^2) = 2ax + 2aky \)
To bring it to the standard form, divide the entire equation by \( \sqrt{1+k^2} \):
\( \implies x^2 + y^2 = \frac{2a}{\sqrt{1+k^2}}x + \frac{2ak}{\sqrt{1+k^2}}y \)
Rearrange the terms to get the general equation of a circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \):
\( \implies x^2 + y^2 - \frac{2a}{\sqrt{1+k^2}}x - \frac{2ak}{\sqrt{1+k^2}}y = 0 \)
By comparing the coefficients, we find:
For x-term: \( 2g = -\frac{2a}{\sqrt{1+k^2}} \implies g = -\frac{a}{\sqrt{1+k^2}} \)
For y-term: \( 2f = -\frac{2ak}{\sqrt{1+k^2}} \implies f = -\frac{ak}{\sqrt{1+k^2}} \)
For the constant term: \( c = 0 \)
The centre \( (-g, -f) \) is calculated as:
\( \left( -\left(-\frac{a}{\sqrt{1+k^2}}\right), -\left(-\frac{ak}{\sqrt{1+k^2}}\right) \right) = \left( \frac{a}{\sqrt{1+k^2}}, \frac{ak}{\sqrt{1+k^2}} \right) \)
The radius \( r \) is found using the formula \( r = \sqrt{g^2 + f^2 - c} \).
\( \implies r = \sqrt{\left(-\frac{a}{\sqrt{1+k^2}}\right)^2 + \left(-\frac{ak}{\sqrt{1+k^2}}\right)^2 - 0} \)
\( \implies r = \sqrt{\frac{a^2}{1+k^2} + \frac{a^2k^2}{1+k^2}} \)
Combine the fractions under the square root:
\( \implies r = \sqrt{\frac{a^2 + a^2k^2}{1+k^2}} \)
Factor out \( a^2 \) from the numerator:
\( \implies r = \sqrt{\frac{a^2(1+k^2)}{1+k^2}} \)
Cancel out \( (1+k^2) \) from numerator and denominator:
\( \implies r = \sqrt{a^2} \)
\( \implies r = a \)
So, the centre is \( \left( \frac{a}{\sqrt{1+k^2}}, \frac{ak}{\sqrt{1+k^2}} \right) \) and the radius is \( a \) units.
(iii) Given the equation of the circle:
\( 4(x^2 + y^2) = 1 \)
To get it into the standard form \( x^2 + y^2 = r^2 \), divide by 4:
\( \implies x^2 + y^2 = \frac{1}{4} \)
Rewrite the right side as a square of the radius:
\( \implies x^2 + y^2 = \left(\frac{1}{2}\right)^2 \)
Comparing this with the standard equation \( x^2 + y^2 = r^2 \) for a circle centered at the origin, we see that the centre is \( (0, 0) \) and the radius \( r \) is \( \frac{1}{2} \) units.
In simple words: To find the center and radius, rewrite the circle's equation into a standard form. For equations like \( x^2 + y^2 + 2gx + 2fy + c = 0 \), the center is \( (-g, -f) \) and radius is \( \sqrt{g^2 + f^2 - c} \). For circles centered at the origin, it's \( x^2 + y^2 = r^2 \), so the center is \( (0,0) \) and radius is \( r \).

🎯 Exam Tip: Always make sure the coefficients of \( x^2 \) and \( y^2 \) are 1 before comparing with the general equation. If they are not, divide the entire equation by that common coefficient first.

Question 4. Find the equation of the circle which cuts x- axis at a distance +3 from origin and cuts on intercept at y-axis of length 6 units.
Answer:
Let the circle be tangent to the x-axis at point E. The distance from the origin O to E is 3, so E is at \( (3, 0) \).
Since the circle is tangent to the x-axis at \( (3,0) \), the x-coordinate of the center \( C \) must be 3. Let the center be \( C = (3, k) \). The radius \( r \) of the circle will be \( |k| \). Since the diagram shows the circle in the first quadrant, we take \( k=r \). So, the center is \( (3, r) \).
The circle makes an intercept of length 6 units on the y-axis. Let this chord be AB.
The perpendicular distance from the center \( C(3,r) \) to the y-axis (which is \( x=0 \)) is the x-coordinate of C, which is 3. This is \( CD = 3 \).
The perpendicular from the center to a chord bisects the chord. So, \( AD = \frac{AB}{2} = \frac{6}{2} = 3 \).
Now, consider the right-angled triangle \( \triangle ADC \). The radius \( CA \) is the hypotenuse.
Using the Pythagorean theorem: \( CA^2 = AD^2 + CD^2 \)
\( \implies r^2 = 3^2 + 3^2 \)
\( \implies r^2 = 9 + 9 \)
\( \implies r^2 = 18 \)
\( \implies r = \sqrt{18} = 3\sqrt{2} \)
So, the radius of the circle is \( 3\sqrt{2} \) units.
The center of the circle is \( (h, k) = (3, 3\sqrt{2}) \).
The equation of the circle is \( (x - h)^2 + (y - k)^2 = r^2 \).
Substitute the values: \( (x - 3)^2 + (y - 3\sqrt{2})^2 = (3\sqrt{2})^2 \)
\( \implies x^2 - 6x + 9 + y^2 - 2(y)(3\sqrt{2}) + (3\sqrt{2})^2 = 18 \)
\( \implies x^2 - 6x + 9 + y^2 - 6\sqrt{2}y + 18 = 18 \)
\( \implies x^2 + y^2 - 6x - 6\sqrt{2}y + 9 + 18 - 18 = 0 \)
\( \implies x^2 + y^2 - 6x - 6\sqrt{2}y + 9 = 0 \)
This is the equation of the circle in the first quadrant.
Similarly, circles can exist in other quadrants with radius \( 3\sqrt{2} \) but with different center coordinates `\( (\pm 3, \pm 3\sqrt{2}) \)`. The possible equations are:
1. For center \( (3, 3\sqrt{2}) \): \( x^2 + y^2 - 6x - 6\sqrt{2}y + 9 = 0 \)
2. For center \( (-3, 3\sqrt{2}) \): \( x^2 + y^2 + 6x - 6\sqrt{2}y + 9 = 0 \)
3. For center \( (-3, -3\sqrt{2}) \): \( x^2 + y^2 + 6x + 6\sqrt{2}y + 9 = 0 \)
4. For center \( (3, -3\sqrt{2}) \): \( x^2 + y^2 - 6x + 6\sqrt{2}y + 9 = 0 \)

In simple words: First, find the radius using the given intercept lengths and tangency information with the Pythagorean theorem. Then, determine the possible center points. Finally, write the circle's equation using the center and radius.

🎯 Exam Tip: When a circle is tangent to an axis, the radius equals the absolute value of the coordinate perpendicular to that axis (e.g., if tangent to x-axis, \( r = |k| \)). Also, remember that a perpendicular from the center bisects a chord.

Question 5. Find the centre and radius of circle \( x^2 + y^2 - 8x + 10y - 12 = 0 \).
Answer:
Given the equation of the circle:
\( x^2 + y^2 - 8x + 10y - 12 = 0 \)
To find the centre and radius, we will complete the square for the x and y terms. First, move the constant term to the right side of the equation:
\( \implies x^2 - 8x + y^2 + 10y = 12 \)
Now, complete the square for the x-terms and y-terms separately.
For x-terms: Take half of the coefficient of x (\(-8\)), square it (\((-4)^2 = 16\)), and add it to both sides.
For y-terms: Take half of the coefficient of y (\(10\)), square it (\((5)^2 = 25\)), and add it to both sides.
\( \implies (x^2 - 8x + 16) + (y^2 + 10y + 25) = 12 + 16 + 25 \)
Rewrite the expressions in parentheses as squared terms:
\( \implies (x - 4)^2 + (y + 5)^2 = 53 \)
To match the standard form \( (x - h)^2 + (y - k)^2 = r^2 \), we rewrite \( (y+5)^2 \) as \( (y - (-5))^2 \) and \( 53 \) as \( (\sqrt{53})^2 \):
\( \implies (x - 4)^2 + (y - (-5))^2 = (\sqrt{53})^2 \)
By comparing this equation with the standard form, we identify the values of \( h \), \( k \), and \( r \):
\( h = 4 \)
\( k = -5 \)
\( r = \sqrt{53} \)
Therefore, the centre of the circle is \( (4, -5) \) and its radius is \( \sqrt{53} \) units.
In simple words: To find the center and radius from an equation, rearrange it to the standard form by completing the square for the x and y parts. The center will be \( (h,k) \) and the radius will be \( r \).

🎯 Exam Tip: Completing the square is a key technique for finding the center and radius of a circle from its general equation. Remember to add the same values to both sides of the equation to maintain balance.

Question 6. Find the centre and radius of the circle \( 2x^2 + 2y^2 - x = 0 \).
Answer:
Given the equation of the circle:
\( 2x^2 + 2y^2 - x = 0 \)
First, to convert the equation to the standard form `\( x^2 + y^2 + 2gx + 2fy + c = 0 \)`, we need the coefficients of \( x^2 \) and \( y^2 \) to be 1. So, divide the entire equation by 2:
\( \implies x^2 + y^2 - \frac{1}{2}x = 0 \)
Now, we rearrange the terms and complete the square for the x-terms. Since there is no linear y-term, we can write \( y^2 \) as \( (y-0)^2 \).
\( \implies \left(x^2 - \frac{1}{2}x\right) + y^2 = 0 \)
To complete the square for \( x^2 - \frac{1}{2}x \), we take half of the coefficient of x (\(-\frac{1}{2}\)), square it (\(\left(-\frac{1}{4}\right)^2 = \frac{1}{16}\)), and add it to both sides of the equation:
\( \implies \left(x^2 - \frac{1}{2}x + \frac{1}{16}\right) + (y - 0)^2 = \frac{1}{16} \)
Rewrite the x-terms as a squared binomial:
\( \implies \left(x - \frac{1}{4}\right)^2 + (y - 0)^2 = \left(\frac{1}{4}\right)^2 \)
We compare this with the standard equation of a circle \( (x - h)^2 + (y - k)^2 = r^2 \).
From this comparison, we identify the values of \( h \), \( k \), and \( r \):
\( h = \frac{1}{4} \)
\( k = 0 \)
\( r = \frac{1}{4} \)
Therefore, the centre of the circle is \( \left(\frac{1}{4}, 0\right) \) and its radius is \( \frac{1}{4} \) units.
In simple words: First, make sure the \( x^2 \) and \( y^2 \) terms have a coefficient of 1. Then, complete the square for the x and y terms to get the equation in the standard form \( (x-h)^2 + (y-k)^2 = r^2 \). From this, you can directly read the center and radius.

🎯 Exam Tip: Always begin by normalizing the equation so that the coefficients of \( x^2 \) and \( y^2 \) are both 1. This prevents errors when identifying \( g, f, c \) or completing the square.

Question 7. Find the equation of the circle passing through the points (2, 3) and (-1, 1) and whose centre lie on line \( x - 3y - 11 = 0 \).
Answer:
Let the general equation of the circle be \( (x - h)^2 + (y - k)^2 = r^2 \).
The problem states that the circle passes through two points: \( (2, 3) \) and \( (-1, 1) \). We substitute these coordinates into the circle's equation.
For point \( (2, 3) \), we get:
\( (2 - h)^2 + (3 - k)^2 = r^2 \)
Expand this equation:
\( \implies 4 - 4h + h^2 + 9 - 6k + k^2 = r^2 \)
\( \implies h^2 + k^2 - 4h - 6k + 13 = r^2 \) (Equation 1)
For point \( (-1, 1) \), we get:
\( (-1 - h)^2 + (1 - k)^2 = r^2 \)
Expand this equation. Note that \( (-1 - h)^2 = (h+1)^2 \):
\( \implies (h+1)^2 + (1 - k)^2 = r^2 \)
\( \implies h^2 + 2h + 1 + 1 - 2k + k^2 = r^2 \)
\( \implies h^2 + k^2 + 2h - 2k + 2 = r^2 \) (Equation 2)
The problem also states that the centre \( (h, k) \) lies on the line \( x - 3y - 11 = 0 \).
Substituting \( h \) for \( x \) and \( k \) for \( y \) into the line's equation:
\( h - 3k - 11 = 0 \)
\( \implies h - 3k = 11 \) (Equation 3)
Now, we have two expressions for \( r^2 \) from Equation 1 and Equation 2. We can set them equal to each other to eliminate \( r^2 \):
\( h^2 + k^2 - 4h - 6k + 13 = h^2 + k^2 + 2h - 2k + 2 \)
Simplify by canceling \( h^2 \) and \( k^2 \) from both sides:
\( \implies -4h - 6k + 13 = 2h - 2k + 2 \)
Rearrange the terms to group \( h \) and \( k \) on one side and constants on the other:
\( \implies 13 - 2 = 2h + 4h - 2k + 6k \)
\( \implies 11 = 6h + 4k \) (Equation 4)
We now have a system of two linear equations for \( h \) and \( k \):
Equation 3: \( h - 3k = 11 \)
Equation 4: \( 6h + 4k = 11 \)
To solve for \( h \) and \( k \), we can use the elimination method. Multiply Equation 3 by 4 and Equation 4 by 3, then add the resulting equations:
Multiply Eq 3 by 4: \( 4(h - 3k) = 4(11) \implies 4h - 12k = 44 \)
Multiply Eq 4 by 3: \( 3(6h + 4k) = 3(11) \implies 18h + 12k = 33 \)
Add these two new equations:
\( (4h - 12k) + (18h + 12k) = 44 + 33 \)
\( \implies 22h = 77 \)
Solve for \( h \):
\( \implies h = \frac{77}{22} \)
\( \implies h = \frac{7}{2} \)
Now, substitute the value of \( h \) into Equation 3 (`\( h - 3k = 11 \)`):
\( \frac{7}{2} - 3k = 11 \)
Solve for \( k \):
\( \implies -3k = 11 - \frac{7}{2} \)
\( \implies -3k = \frac{22 - 7}{2} \)
\( \implies -3k = \frac{15}{2} \)
\( \implies k = \frac{15}{2 \times (-3)} \)
\( \implies k = -\frac{15}{6} \)
\( \implies k = -\frac{5}{2} \)
So, the centre of the circle is \( (h, k) = \left(\frac{7}{2}, -\frac{5}{2}\right) \).
Now, substitute the values of \( h \) and \( k \) into Equation 1 (`\( h^2 + k^2 - 4h - 6k + 13 = r^2 \)` ) to find \( r^2 \):
\( \left(\frac{7}{2}\right)^2 + \left(-\frac{5}{2}\right)^2 - 4\left(\frac{7}{2}\right) - 6\left(-\frac{5}{2}\right) + 13 = r^2 \)
\( \implies \frac{49}{4} + \frac{25}{4} - 14 + 15 + 13 = r^2 \)
\( \implies \frac{74}{4} - 14 + 15 + 13 = r^2 \)
\( \implies \frac{37}{2} + 14 = r^2 \)
\( \implies \frac{37}{2} + \frac{28}{2} = r^2 \)
\( \implies r^2 = \frac{37 + 28}{2} \)
\( \implies r^2 = \frac{65}{2} \)
Therefore, the equation of the circle \( (x - h)^2 + (y - k)^2 = r^2 \) becomes:
\( \left(x - \frac{7}{2}\right)^2 + \left(y - \left(-\frac{5}{2}\right)\right)^2 = \frac{65}{2} \)
\( \implies \left(x - \frac{7}{2}\right)^2 + \left(y + \frac{5}{2}\right)^2 = \frac{65}{2} \)
Expand the squared terms:
\( \implies x^2 - 2x\left(\frac{7}{2}\right) + \left(\frac{7}{2}\right)^2 + y^2 + 2y\left(\frac{5}{2}\right) + \left(\frac{5}{2}\right)^2 = \frac{65}{2} \)
\( \implies x^2 - 7x + \frac{49}{4} + y^2 + 5y + \frac{25}{4} = \frac{65}{2} \)
Combine constant terms:
\( \implies x^2 + y^2 - 7x + 5y + \frac{49+25}{4} = \frac{65}{2} \)
\( \implies x^2 + y^2 - 7x + 5y + \frac{74}{4} = \frac{65}{2} \)
\( \implies x^2 + y^2 - 7x + 5y + \frac{37}{2} = \frac{65}{2} \)
Move all terms to one side:
\( \implies x^2 + y^2 - 7x + 5y + \frac{37}{2} - \frac{65}{2} = 0 \)
\( \implies x^2 + y^2 - 7x + 5y - \frac{28}{2} = 0 \)
\( \implies x^2 + y^2 - 7x + 5y - 14 = 0 \)
This is the required equation of the circle.
In simple words: When a circle passes through points and its center is on a line, use the general circle equation. Substitute the points to get equations for radius squared. Use the line equation for the center's coordinates. Solve the system of equations to find the center and radius, then write the final circle equation.

🎯 Exam Tip: This type of problem requires solving a system of equations. Ensure your algebraic manipulations are accurate, especially when expanding and combining terms. Double-check your signs throughout the calculation.

Question 8. Find the equation of a circle with radius 5, whose center lies on the x-axis, and which passes through the point (2, 3).
Answer:
Let the centre of the circle be \( (h, k) \). Since the centre lies on the x-axis, its y-coordinate \( k \) must be 0.
The radius is given as \( r = 5 \).
The general equation of the circle is \( (x - h)^2 + (y - k)^2 = r^2 \).
Substituting \( k = 0 \) and \( r = 5 \) into the equation, we get:
\( (x - h)^2 + (y - 0)^2 = 5^2 \)
\( \implies (x - h)^2 + y^2 = 25 \) (Equation 1)
The problem states that the circle passes through the point \( (2, 3) \). We substitute these coordinates into Equation 1:
\( (2 - h)^2 + 3^2 = 25 \)
\( \implies (2 - h)^2 + 9 = 25 \)
\( \implies (2 - h)^2 = 25 - 9 \)
\( \implies (2 - h)^2 = 16 \)
Now, take the square root of both sides:
\( \implies 2 - h = \pm 4 \)
This gives us two possible values for \( h \):
Case 1: \( 2 - h = 4 \)
\( \implies h = 2 - 4 \)
\( \implies h = -2 \)
Case 2: \( 2 - h = -4 \)
\( \implies h = 2 + 4 \)
\( \implies h = 6 \)
So, there are two possible centres for the circle: \( (-2, 0) \) or \( (6, 0) \).
Now we find the equation for each case.
Case A: Centre \( (6, 0) \) and radius \( r = 5 \).
Using Equation 1: \( (x - 6)^2 + y^2 = 25 \)
Expand the square:
\( \implies x^2 - 12x + 36 + y^2 = 25 \)
Rearrange the terms to get the general equation:
\( \implies x^2 + y^2 - 12x + 36 - 25 = 0 \)
\( \implies x^2 + y^2 - 12x + 11 = 0 \)
Case B: Centre \( (-2, 0) \) and radius \( r = 5 \).
Using Equation 1: \( (x - (-2))^2 + y^2 = 25 \)
\( \implies (x + 2)^2 + y^2 = 25 \)
Expand the square:
\( \implies x^2 + 4x + 4 + y^2 = 25 \)
Rearrange the terms to get the general equation:
\( \implies x^2 + y^2 + 4x + 4 - 25 = 0 \)
\( \implies x^2 + y^2 + 4x - 21 = 0 \)
Thus, there are two possible equations for the circle: \( x^2 + y^2 - 12x + 11 = 0 \) and \( x^2 + y^2 + 4x - 21 = 0 \).
In simple words: When the center is on an axis, one coordinate is zero. Use the given radius and the point the circle passes through to find the missing center coordinate. This often leads to two possible circles.

🎯 Exam Tip: Remember that taking the square root can result in both positive and negative values, leading to multiple possible solutions for the center and thus multiple circle equations.

Question 9. Find the equation of circle that passes through point (0, 0) and cuts intercepts a and b at axis.
Answer:
The problem states that the circle passes through the origin \( (0, 0) \), and cuts intercepts \( a \) and \( b \) on the axes. This means it also passes through points \( (a, 0) \) (on the x-axis) and \( (0, b) \) (on the y-axis).
Let the general equation of the circle be \( (x - h)^2 + (y - k)^2 = r^2 \).
Since the circle passes through these three points, we can substitute each point's coordinates into the equation to form a system of equations.
1. Substitute the point \( (0, 0) \) into the equation:
\( (0 - h)^2 + (0 - k)^2 = r^2 \)
\( \implies h^2 + k^2 = r^2 \) (Equation 1)
2. Substitute the point \( (a, 0) \) into the equation:
\( (a - h)^2 + (0 - k)^2 = r^2 \)
\( \implies (a - h)^2 + k^2 = r^2 \) (Equation 2)
3. Substitute the point \( (0, b) \) into the equation:
\( (0 - h)^2 + (b - k)^2 = r^2 \)
\( \implies h^2 + (b - k)^2 = r^2 \) (Equation 3)
From Equation 1 and Equation 2, we can equate the expressions for \( r^2 \):
\( h^2 + k^2 = (a - h)^2 + k^2 \)
Cancel \( k^2 \) from both sides:
\( \implies h^2 = (a - h)^2 \)
Expand the right side:
\( \implies h^2 = a^2 - 2ah + h^2 \)
Cancel \( h^2 \) from both sides:
\( \implies 0 = a^2 - 2ah \)
\( \implies 2ah = a^2 \)
Assuming \( a \neq 0 \) (since it's an intercept), we can divide by \( a \):
\( \implies 2h = a \)
\( \implies h = \frac{a}{2} \)
Similarly, from Equation 1 and Equation 3, we can equate the expressions for \( r^2 \):
\( h^2 + k^2 = h^2 + (b - k)^2 \)
Cancel \( h^2 \) from both sides:
\( \implies k^2 = (b - k)^2 \)
Expand the right side:
\( \implies k^2 = b^2 - 2bk + k^2 \)
Cancel \( k^2 \) from both sides:
\( \implies 0 = b^2 - 2bk \)
\( \implies 2bk = b^2 \)
Assuming \( b \neq 0 \) (since it's an intercept), we can divide by \( b \):
\( \implies 2k = b \)
\( \implies k = \frac{b}{2} \)
So, the coordinates of the centre are \( (h, k) = \left(\frac{a}{2}, \frac{b}{2}\right) \).
Now, substitute the values of \( h \) and \( k \) into Equation 1 (`\( h^2 + k^2 = r^2 \)` ) to find \( r^2 \):
\( r^2 = \left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2 \)
\( \implies r^2 = \frac{a^2}{4} + \frac{b^2}{4} \)
\( \implies r^2 = \frac{a^2 + b^2}{4} \)
Finally, substitute the values of \( h \), \( k \), and \( r^2 \) back into the general equation of the circle \( (x - h)^2 + (y - k)^2 = r^2 \):
\( \left(x - \frac{a}{2}\right)^2 + \left(y - \frac{b}{2}\right)^2 = \frac{a^2 + b^2}{4} \)
Expand the squared terms on the left side:
\( \implies x^2 - 2x\left(\frac{a}{2}\right) + \left(\frac{a}{2}\right)^2 + y^2 - 2y\left(\frac{b}{2}\right) + \left(\frac{b}{2}\right)^2 = \frac{a^2 + b^2}{4} \)
\( \implies x^2 - ax + \frac{a^2}{4} + y^2 - by + \frac{b^2}{4} = \frac{a^2 + b^2}{4} \)
Group the terms:
\( \implies x^2 + y^2 - ax - by + \frac{a^2}{4} + \frac{b^2}{4} = \frac{a^2 + b^2}{4} \)
\( \implies x^2 + y^2 - ax - by + \frac{a^2 + b^2}{4} = \frac{a^2 + b^2}{4} \)
Subtract \( \frac{a^2 + b^2}{4} \) from both sides of the equation:
\( \implies x^2 + y^2 - ax - by = 0 \)
This is the required equation of the circle that passes through the origin and makes intercepts \( a \) and \( b \) on the axes.
In simple words: A circle passing through the origin and making intercepts 'a' and 'b' on the axes means it goes through \( (0,0) \), \( (a,0) \), and \( (0,b) \). Use these three points with the standard circle equation to find the center and radius, which then gives you the final equation.

🎯 Exam Tip: When a problem states intercepts, translate them into coordinate points (e.g., x-intercept 'a' means point \( (a,0) \)). Using three points to find the circle's equation often involves setting up and solving simultaneous equations.