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      <text>2.6.1.5 Example: distance between any 2 points</text>
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<h4 class="HELM_example">Example</h4>
<p>Generalise your result to the previous Task to obtain the distance between any two points whose coordinates are \((x_1, y_1)\) and \((x_2, y_2).\)</p>

<h4 class="HELM_solution">Solution</h4>

<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/circle4a.svg" alt="Image showing the distance between 2 points, P1 (x1,y1) and P2(x2,y2), on the xy coordinate system. " width="337" height="168" class="img-responsive atto_image_button_text-bottom">
</figure>

<p>Between the arbitrary points \(P_1\) and \(P_2\) the distance is: \[d= \sqrt{(AP_2)^2 + (P_1A)^2} \]</p>
<p>where \(AP_2 = x_2-x_1, \, P_1A = y_1-y_2 = -(y_2-y_1), \) so \[d= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]</p>

<hr>
<p>Using the result of the last Task, we now consider a circle centre at the point \(C \, (x_0, y_0)\) and of radius \(R.\) Suppose P is an arbitrary point on this circle which has co-ordinates \((x, y)\):</p>

<figure class="HELM_figure">
  <img src="@@PLUGINFILE@@/circle5.svg" alt="A circle of radius R, centered at the point C(x0,y0), and a random point P(x,y) on the circle. " width="160" height="127" class="img-responsive atto_image_button_text-bottom">
</figure>

<p>Clearly \( R=CP= \sqrt{(x-x_0)^2 = (y-y_0)^2}\)</p>
<p>Hence, squaring both sides,</p>
<p>
</p>
<p>\( \qquad (x-x_0)^2 + (y-y_0)^2 = R^2 \qquad\) (2)</p>
<p>which is said to be the equation of the circle centre \((x_0, y_0)\) radius \(R.\)</p>
<p>Note that if \(x_0 = y_0 = 0\) (i.e. circle centre is at origin) then (2) reduces to (1) so the latter is simply a special case.</p>
<p>The interpretation of (2) is similar to that of (1): any point \((x, y)\) satisfying (2) lies on the circumference of the circle.</p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>What does the equation \( (x-3)^2 + (y-4)^2 = 4\) present?</p>

<h4 class="HELM_solution">Solution</h4>
<p>It represents a circle of radius \(2\) (the positive square root of \(4\)) and has centre \(C \, (3, 4).\)</p>
<p>N.B. There is no need to expand the terms on the left-hand side of the equation here. The given form reveals quite plainly the radius (\(\sqrt{4}\)) and centre \((3, 4)\) of the circle.</p>]]></text>
<file name="circle4a.svg" path="/" encoding="base64">[encoded data, 36716 characters]</file>
<file name="circle5.svg" path="/" encoding="base64">[encoded data, 25012 characters]</file>
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