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    <p>Color.BLACK is the integer representation of argb Hex value 0xff000000. So your statement is checking whether a center point in the circle is exactly the same transparency, red value, blue value and green value as Color.BLACK.</p> <p>A few options:</p> <p>You can try comparing just the rgb value by using </p> <pre><code>if(Color.rgb(Color.red(Color.BLACK), Color.green(Color.BLACK), Color.blue(Color.BLACK) == Color.rgb(Color.red(pixelColor), Color.green(pixelColor), Color.blue(pixelColor)) </code></pre> <p>Alternatively you could scan the entire circle for a black (0x000000) pixel.</p> <p>Alternatively you could use a Color difference algorithm and you can test different tolerances for what you need. This may help you <a href="https://stackoverflow.com/questions/9018016/how-to-compare-two-colors">How to compare two colors for similarity/difference</a>.</p> <p>The following hasn't been tested, but will give you an idea of which direction you could take also: </p> <pre><code>//mid points x1 and y1 int x1=(int) (myMidPoint.x) /2; int y1=(int)(myMidPoint.y)/2; int radius = 30; int topPoint = y1 - radius; int leftPoint = x1 - radius; int rightPoint = x1 + radius; int bottomPoint = y1 + radius; int scanWidth = 0; for(int i = topPoint; i &lt; bottomPoint; i++) { if(i &lt;= y1) { scanWidth++; } else { scanWidth--; } for(int j = x1 - scanWidth; j &lt; x1 + scanWidth; j++) { int pixelColor = myBitmap.getPixel(j,i); if(Color.rgb(Color.red(Color.BLACK), Color.green(Color.BLACK), Color.blue(Color.BLACK) == Color.rgb(Color.red(pixelColor), Color.green(pixelColor), Color.blue(pixelColor)) { System.out.println("pixel black"); } } } </code></pre>
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    1. PODetecting black pixel in an image
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    1. COif i want to scan the entire circle for a black pixel what should i do..how i can access the circle which is drawn by canvas.draw circle
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    2. COcomparing rgb value not works well..Any suggestions to scan the entire circle for a black pixel
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    3. COWhat efficiency are you looking to achieve? I can try and chop out some code for you to try, but it won't be 100% efficient
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