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The circle with equation $x^2 + y^2 - 12x - 10y + k = 0$ meets the coordinate axes at exactly three points - Scottish Highers Maths - Question 14 - 2015

Question 14

The circle with equation $x^2 + y^2 - 12x - 10y + k = 0$ meets the coordinate axes at exactly three points. What is the value of k?

Worked Solution & Example Answer:The circle with equation $x^2 + y^2 - 12x - 10y + k = 0$ meets the coordinate axes at exactly three points - Scottish Highers Maths - Question 14 - 2015

Step 1

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Answer

To find the radius of the circle, we first rewrite the equation in standard form by completing the square.

The given equation is: $x^2 - 12x + y^2 - 10y + k = 0$

Completing the square for the $x$ terms: $x^2 - 12x = (x - 6)^2 - 36$

Completing the square for the $y$ terms: $y^2 - 10y = (y - 5)^2 - 25$

So we rewrite the equation as: $(x - 6)^2 + (y - 5)^2 = 36 + 25 - k$

This leads to: $(x - 6)^2 + (y - 5)^2 = 61 - k$

Hence, the radius $r$ of the circle is given by: $r = ext{sqrt}(61 - k)$

Step 2

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Answer

For the circle to meet the coordinate axes at exactly three points, it must be tangent to one axis and intersect the other.

The circle touches the $y$ -axis at one point. This occurs when: $r = 6$ Therefore: $ext{sqrt}(61 - k) = 6$ Squaring both sides gives: $61 - k = 36$ Thus: $k = 25$
The circle passes through the origin (0, 0) implying: $r = 0$ This condition also leads to: $61 - k = 0$ Hence: $k = 61$

In order to meet the required condition, we find the only appropriate value is: $k = 25$

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