Question:
<p>Let the parabola <span class="math-tex">\(y=x^{2}+p x-3\)</span>, meet the coordinate axes at the points <span class="math-tex">\({P}, {Q}\)</span> and R. If the circle C with centre at <span class="math-tex">\((-1,-1)\)</span> passes through the points <span class="math-tex">\({P}, {Q}\)</span> and R, then the area of <span class="math-tex">\(\triangle {PQR}\)</span> is:</p>
<p style="display:inline">6</p>
<p style="display:inline">5</p>
<p style="display:inline">4</p>
<p style="display:inline">7</p>
Step-by-Step Solution
Key Concept: Determine the circle's equation using its center and the parabola's y-intercept, then find the circle's x-intercepts to identify the remaining vertices of the triangle.
<p><span class="math-tex">$y=x^{2}+p x-3$</span><br />
<span class="math-tex">$\because x=0 y=-3$</span><br />
<span class="math-tex">$\therefore(0,-3)$</span> be point where parabola (cuts <span class="math-tex">$y$</span> axis)<br />
<span class="math-tex">$\therefore$</span> Equation of circle 'C'<br />
<span class="math-tex">$(x+1)^{2}+(y+1)^{2}=(-1-0)^{2}+(-1+3)^{2}$</span><br />
<span class="math-tex">$\Rightarrow(x+1)^{2}+(y+1)^{2}=5$</span><br />
<img src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/1757581545-s7q8dp.jpg" style="height:118px; width:250px" /><br />
Circle cuts <span class="math-tex">$x$</span>-axis at <span class="math-tex">$y=0$</span><br />
<span class="math-tex">$\therefore(x+1)^{2}=4 \Rightarrow x+1$</span> <span class="math-tex">$= \pm 2 \Rightarrow x=1,-3$</span><br />
<span class="math-tex">$\therefore$</span> Points are <span class="math-tex">$(1,0),(-3,0)$</span> and <span class="math-tex">$(0,-3)$</span><br />
Area of <span class="math-tex">$\Delta=\frac{1}{2} \times 4 \times 3=6$</span></p>
Correct Answer: A