Question:
<p>Find the equation of circle which pass through (5, 9) and center at (2, 5).</p>
<p style="display:inline"><span class="math-tex">\(x^2+y^2+4 x-10 y-4=0\)</span></p>
<p style="display:inline"><span class="math-tex">\(x^2+y^2-4 x-10 y+4=0\)</span></p>
<p style="display:inline">a) <span class="math-tex">\(x^2+y^2+4 x-10 y+4=0\)</span></p>
<p style="display:inline"><span class="math-tex">\(x^2+y^2+4 x+10 y+4=0\)</span></p>
Step-by-Step Solution
Key Concept: The equation of a circle is uniquely determined by its center and its radius, where the radius is the distance from the center to any point on the circle's circumference.
<p>Equation of circle with center at <span class="math-tex">$(a, b)$</span> and radius <span class="math-tex">$r$</span> units is<br />
<span class="math-tex">$ (x-a)^2+(y-b)^2=r^2 $</span><br />
<span class="math-tex">$ (5-2)^2+(9-5)^2=r^2=>r^2=3^2+4^2=>r=5 . $</span><br />
So, equation of circle is <span class="math-tex">$(x-2)^2+(y-5)^2=5^2=>x^2+y^2-4 x-10 y+4=0$</span>.</p>
Correct Answer: B