Question Statement

Check whether the given point lies inside, outside, or on the given circle:

(i) $(3,5)$ ; $x^2+y^2-6x+2y-18=0$

(ii) $(1,6)$ ; $2x^2+2y^2-4x-16y+5=0$

(iii) $(-1,-2)$ ; $x^2+y^2+6x+4y+9=0$

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Background and Explanation

To determine the position of a point relative to a circle, substitute the point coordinates $(x_1,y_1)$ into the left-hand side of the circle equation $S=x^2+y^2+2gx+2fy+c$. If the result is positive, the point lies outside; if negative, inside; if zero, on the circle. This method works because the expression represents the power of the point with respect to the circle.

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Solution

Part (i)

Given: Point $(x_1,y_1)=(3,5)$ and circle $x^2+y^2-6x+2y-18=0$

Substitute the coordinates $(3,5)$ into the left-hand side of the circle equation:

$\begin{array}{rl}{x}_1^2+y_1^2-6x_1+2y_1-18& =(3{)}^2+(5{)}^2-6(3)+2(5)-18\\ & =9+25-18+10-18\\ & =8\end{array}$

Since the result is positive ($8>0$), the point lies outside the circle.

Part (ii)

Given: Point $(x_1,y_1)=(1,6)$ and circle $2x^2+2y^2-4x-16y+5=0$

First, make the coefficients of $x^2$ and $y^2$ equal to $1$ by dividing the entire equation by $2$:

$x^2+y^2-2x-8y+\frac{5}{2}=0$

Now substitute the point $(1,6)$ into the normalized equation:

$\begin{array}{rl}{x}_1^2+y_1^2-2x_1-8y_1+\frac{5}{2}& =(1{)}^2+(6{)}^2-2(1)-8(6)+\frac{5}{2}\\ & =1+36-2-48+\frac{5}{2}\\ & =-\frac{21}{2}\end{array}$

Since the result is negative ($-\frac{21}{2}<0$), the point lies inside the circle.

Part (iii)

Given: Point $(x_1,y_1)=(-1,-2)$ and circle $x^2+y^2+6x+4y+9=0$

Substitute the coordinates $(-1,-2)$ into the left-hand side of the circle equation:

$\begin{array}{rl}{x}_1^2+y_1^2+6x_1+4y_1+9& =(-1{)}^2+(-2{)}^2+6(-1)+4(-2)+9\\ & =1+4-6-8+9\\ & =0\end{array}$

Since the result equals zero, the point lies on the circle.

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Key Formulas or Methods Used

• Position of point relative to circle: For a circle $x^2+y^2+2gx+2fy+c=0$ and point $(x_1,y_1)$, calculate: $S_1=x_1^2+y_1^2+2g{x}_1+2f{y}_1+c$

  • If $S_1>0$: Point lies outside the circle
  • If $S_1<0$: Point lies inside the circle
  • If $S_1=0$: Point lies on the circle

• Normalization of circle equation: When coefficients of $x^2$ and $y^2$ are not $1$ (but are equal), divide the entire equation by that coefficient before substituting the point.

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Summary of Steps

1. Identify the given point $(x_1,y_1)$ and the circle equation.
2. Normalize the circle equation if needed: divide by the coefficient of $x^2$ (when it equals the coefficient of $y^2$) to make leading coefficients equal to $1$.
3. Substitute $x_1$ for $x$ and $y_1$ for $y$ in the left-hand side of the circle equation.
4. Evaluate the expression and determine position:
   • Positive $\Rightarrow$ Outside
   • Negative $\Rightarrow$ Inside
   • Zero $\Rightarrow$ On the circle