Question Statement

Find the equation of a circle when its centre and radius are given:

(i) Centre at $(3,-1)$ and radius $2$

(ii) Centre at $\left(-\frac{1}{2},-\frac{1}{3}\right)$ and radius $\frac{1}{5}$

(iii) Centre at $\left(\frac{1}{a},a\right)$ and radius $a$ where $a\ne 0$

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Background and Explanation

The standard equation of a circle with center $(h,k)$ and radius $r$ is $(x-h{)}^2+(y-k{)}^2=r^2$. To find the specific equation, substitute the given center coordinates and radius, then expand and simplify to obtain the general form $x^2+y^2+Dx+Ey+F=0$.

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Solution

Part (i): Centre at $(3,-1)$ and radius $2$

The standard equation of a circle is:

$(x-h{)}^2+(y-k{)}^2=r^2$

Substituting $h=3$, $k=-1$, and $r=2$:

$(x-3{)}^2+(y-(-1){)}^2=2^2$

Simplifying the expression:

$(x-3{)}^2+(y+1{)}^2=4$

Expanding the squared binomials:

$x^2-6x+9+y^2+2y+1=4$

Combining like terms and bringing all terms to the left side:

$x^2+y^2-6x+2y+9+1-4=0$

$x^2+y^2-6x+2y+6=0$

Part (ii): Centre at $\left(-\frac{1}{2},-\frac{1}{3}\right)$ and radius $\frac{1}{5}$

Using the standard equation with $h=-\frac{1}{2}$, $k=-\frac{1}{3}$, and $r=\frac{1}{5}$:

${\left(x-\left(-\frac{1}{2}\right)\right)}^2+{\left(y-\left(-\frac{1}{3}\right)\right)}^2={\left(\frac{1}{5}\right)}^2$

Simplifying the double negatives:

${\left(x+\frac{1}{2}\right)}^2+{\left(y+\frac{1}{3}\right)}^2=\frac{1}{25}$

Expanding using the formula $(a+b{)}^2=a^2+2ab+b^2$:

$x^2+x+\frac{1}{4}+y^2+\frac{2}{3}y+\frac{1}{9}=\frac{1}{25}$

Rearranging terms to combine constants:

$x^2+y^2+x+\frac{2}{3}y+\frac{1}{4}+\frac{1}{9}-\frac{1}{25}=0$

Finding a common denominator (900) for the constant terms:

$\frac{225}{900}+\frac{100}{900}-\frac{36}{900}=\frac{289}{900}$

Thus:

$x^2+y^2+x+\frac{2}{3}y+\frac{289}{900}=0$

Multiplying through by 900 to clear all denominators:

$900x^2+900y^2+900x+600y+289=0$

Part (iii): Centre at $\left(\frac{1}{a},a\right)$ and radius $a$ where $a\ne 0$

Using the standard equation with $h=\frac{1}{a}$, $k=a$, and $r=a$:

${\left(x-\frac{1}{a}\right)}^2+(y-a{)}^2=a^2$

Expanding both squared terms:

$x^2-\frac{2}{a}x+\frac{1}{a^2}+y^2-2ay+a^2=a^2$

Subtracting $a^2$ from both sides:

$x^2+y^2-\frac{2}{a}x-2ay+\frac{1}{a^2}=0$

Multiplying through by $a^2$ to eliminate the fraction:

$a^2{x}^2+a^2{y}^2-2ax-2a^{3}y+1=0$

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Key Formulas or Methods Used

• Standard equation of a circle: $(x-h{)}^2+(y-k{)}^2=r^2$ where $(h,k)$ is the center and $r$ is the radius
• Binomial expansion: $(a+b{)}^2=a^2+2ab+b^2$ and $(a-b{)}^2=a^2-2ab+b^2$
• Clearing fractions: Multiplying the entire equation by the least common denominator to obtain integer coefficients

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Summary of Steps

1. Identify parameters: Extract the center coordinates $(h,k)$ and radius $r$ from the given information
2. Substitute: Plug these values into the standard circle equation $(x-h{)}^2+(y-k{)}^2=r^2$
3. Expand: Use binomial formulas to expand the squared terms and remove parentheses
4. Simplify: Combine like terms and move all terms to the left side to get the general form $x^2+y^2+Dx+Ey+F=0$
5. Rationalize (if needed): Multiply through by the appropriate factor to eliminate fractions and obtain integer coefficients