IPMAT Indore 2024 · Quantitative Ability · Question 26

Step 1 — find the centre and radius. For {x}^{2}+{y}^{2}+2gx+2fy+c=0, the centre is \left( -g,-f \right) and r=\sqrt{{g}^{2}+{f}^{2}-c}. Here 2g=-14, 2f=16, c=88:

\text{Centre}C\left( 7,-8 \right), r=\sqrt{49+64-88}=\sqrt{25}=5

Step 2 — distance from A to the centre.

{d}^{2}={\left( 0-7 \right)}^{2}+{\left( 7/2+8 \right)}^{2}=49+{\left( 23/2 \right)}^{2}=49+\frac{529}{4}=\frac{725}{4}

Step 3 — half-angle of the tangent pair. The tangent length is \sqrt{{d}^{2}-{r}^{2}} and

\tan \frac{\theta}{2}=\frac{r}{\sqrt{{d}^{2}-{r}^{2}}}=\frac{5}{\sqrt{\frac{725}{4}-25}}=\frac{5}{\sqrt{\frac{625}{4}}}=\frac{5}{\frac{25}{2}}=\frac{2}{5}

Step 4 — double the angle. With t=\tan \theta /2=2/5:

\tan \theta =\frac{2t}{1-{t}^{2}}=\frac{\frac{4}{5}}{1-\frac{4}{25}}=\frac{\frac{4}{5}}{\frac{21}{25}}=\frac{4}{5}\cdot \frac{25}{21}=\frac{20}{21}

Answer: Option 2 — \frac{20}{21}

Shortcut. Two formulas, in this order.

\tan \frac{\theta}{2}=\frac{r}{L}\text{where}L=\sqrt{{d}^{2}-{r}^{2}}\text{is the tangent length;}\tan \theta =\frac{2t}{1-{t}^{2}}

The numbers are engineered to be clean: {d}^{2}-{r}^{2}=625/4, so L=25/2 and t=2/5 exactly.

The trap is options (c) and (d): 2/5 is \tan \theta /2 and 4/5 is 2t — both are stages on the way, not the answer. Whenever a tangent-pair question gives you a neat half-angle, the printed options will include it.