# Find the equation to the circles which pass through the origin and cut off intercepts equal to

(i) $3$ and $4,$

(ii) $2a$ and $2b$ from the x-axis and the y-axis respectively.

Answer

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**Hint**: Here we will use the standard equation of the circle and then place the coordinates given and then will simplify the equations for the resultant value.

\[{x^2} + {y^2} + 2gx + 2fy + c = 0\]

**:**

__Complete step-by-step answer__We know that the standard equation of the circle can be given by –

\[{x^2} + {y^2} + 2gx + 2fy + c = 0\] ..... (A)

Case (i) Circle passes through the origin and cut off intercepts equal to $3$ and $4,$from the x-axis and the y-axis respectively.

So, the co-ordinates will be $(0,0),(3,0)$ and $(0,4)$

First of all place $(x,y) = (0,0)$ in the equation (A)

\[{(0)^2} + {(0)^2} + 2g(0) + 2f(0) + c = 0\]

By the property zero multiplied with any number gives zero as the resultant value.

\[ \Rightarrow 0 + 0 + 0 + 0 + c = 0\]

Again, by using the additive identity –

$ \Rightarrow c = 0$ .... (B)

Similarly, place $(x,y) = (3,0)$in the equation (A)

\[ \Rightarrow {(3)^2} + {(0)^2} + 2g(3) + 2f(0) + c = 0\]

Simplify the above equation –

\[ \Rightarrow 9 + 6g + c = 0\]

Also place value in the above equation from the equation (B)

\[ \Rightarrow 9 + 6g = 0\]

Take constants on the other side of the equation. When you move any term from one side to another, then the sign of the term also changes. Positive terms become negative and vice-versa.

\[ \Rightarrow 6g = - 9\]

The above equation can be written as –

\[ \Rightarrow 3 \times 2g = - 9\]

Term multiplicative on one side is moved to the opposite side then it goes to the denominator.

\[ \Rightarrow 2g = \dfrac{{ - 9}}{3}\]

\[ \Rightarrow 2g = \dfrac{{ - 3 \times 3}}{3}\]

Common multiples from the numerator and the denominator cancel each other.

\[ \Rightarrow 2g = ( - 3)\] .... (C)

Similarly place $(x,y) = (0,4)$in the equation (A)

\[ \Rightarrow {(0)^2} + {(4)^2} + 2g(0) + 2f(4) + c = 0\]

Simplify the above equation –

\[ \Rightarrow 16 + 8f + c = 0\]

Also place value in the above equation from the equation (B)

\[ \Rightarrow 16 + 8f = 0\]

Take constants on the other side of the equation. When you move any term from one side to another, then the sign of the term also changes. Positive terms become negative and vice-versa.

\[ \Rightarrow 8f = - 16\]

The above equation can be written as –

\[ \Rightarrow 4 \times 2f = - 16\]

Term multiplicative on one side is moved to the opposite side then it goes to the denominator.

\[ \Rightarrow 2f = \dfrac{{ - 16}}{4}\]

\[ \Rightarrow 2f = \dfrac{{ - 4 \times 4}}{4}\]

Common multiples from the numerator and the denominator cancel each other.

\[ \Rightarrow 2f = ( - 4)\] .... (D)

Now, place values of equation (B), (C), (D) in (A)

\[{x^2} + {y^2} + ( - 3)x + ( - 4)y = 0\]

Simplify the above equation –

\[ \Rightarrow {x^2} + {y^2} - 3x - 4y = 0\]

This is the required answer.

**So, the correct answer is “ \[ {x^2} + {y^2} - 3x - 4y = 0\] ”.**

Similarly,

Case (II) ) Circle passes through the origin and cuts off intercepts equal to $2a$ and $2b$from the x-axis and the y-axis respectively.

So, the co-ordinates will be $(0,0),(2a,0)$ and $(0,2b)$

First of all place $(x,y) = (0,0)$ in the equation (A)

\[{(0)^2} + {(0)^2} + 2g(0) + 2f(0) + c = 0\]

By the property zero multiplied with any number gives zero as the resultant value.

\[ \Rightarrow 0 + 0 + 0 + 0 + c = 0\]

Again, by using the additive identity –

$ \Rightarrow c = 0$ .... (F)

Similarly, place $(x,y) = (2a,0)$in the equation (A)

\[ \Rightarrow {(2a)^2} + {(0)^2} + 2g(2a) + 2f(0) + c = 0\]

Simplify the above equation –

\[ \Rightarrow 4{a^2} + 4ag + c = 0\]

Also place value in the above equation from the equation (B)

\[ \Rightarrow 4{a^2} + 4ag = 0\]

Take constants on the other side of the equation. When you move any term from one side to another, then the sign of the term also changes. Positive terms become negative and vice-versa.

\[ \Rightarrow 4ag = - 4{a^2}\]

The above equation can be written as –

\[ \Rightarrow 2 \times 2ag = - 4{a^2}\]

Term multiplicative on one side is moved to the opposite side then it goes to the denominator.

\[ \Rightarrow 2g = \dfrac{{ - 4{a^2}}}{{2a}}\]

Common multiples from the numerator and the denominator cancel each other.

\[ \Rightarrow 2g = ( - 2a)\] .... (G)

Similarly place $(x,y) = (0,2b)$in the equation (A)

\[ \Rightarrow {(0)^2} + {(2b)^2} + 2g(0) + 2f(2b) + c = 0\]

Simplify the above equation –

\[ \Rightarrow 4{b^2} + 4bf + c = 0\]

Also place value in the above equation from the equation (F)

\[ \Rightarrow 4{b^2} + 4bf = 0\]

Take constants on the other side of the equation. When you move any term from one side to another, then the sign of the term also changes. Positive terms become negative and vice-versa.

\[ \Rightarrow 4bf = - 4{b^2}\]

The above equation can be written as –

\[ \Rightarrow 2 \times 2bf = - 4{b^2}\]

Term multiplicative on one side is moved to opposite side then it goes to the denominator.

\[ \Rightarrow 2f = \dfrac{{ - 4{b^2}}}{{2b}}\]

\[ \Rightarrow 2f = \dfrac{{ - 2b \times 2b}}{{2b}}\]

Common multiples from the numerator and the denominator cancel each other.

\[ \Rightarrow 2f = ( - 2b)\] .... (H)

Now, place values of equation (F), (G), (H) in (A)

\[{x^2} + {y^2} + ( - 2a)x + ( - 2b)y = 0\]

Simplify the above equation –

\[{x^2} + {y^2} - 2ax - 2by = 0\]

This is the required answer.

**So, the correct answer is “ \[{x^2} + {y^2} - 2ax - 2by = 0\] ”.**

**Note**: Be careful while simplifying the equations and remember when you move any term from one side to another, the sign also changes. Positive term becomes negative and negative term becomes positive.

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