IPMAT Indore 2023 · Quantitative Ability · Question 33

Step 1 — find the centre and radius. Comparing with {x}^{2}+{y}^{2}+2gx+2fy+c=0: 2g=-6, 2f=4, c=-12, so

\text{Centre}C\left( 3,-2 \right), r=\sqrt{{g}^{2}+{f}^{2}-c}=\sqrt{9+4+12}=\sqrt{25}=5

Step 2 — impose tangency. All four options use lines 4x+3y+k=0. A line is tangent exactly when its distance from the centre equals the radius:

\frac{\left| 4\left( 3 \right)+3\left( -2 \right)+k \right|}{\sqrt{{4}^{2}+{3}^{2}}}=5

\frac{\left| 12-6+k \right|}{5}=5 \Rightarrow \left| k+6 \right|=25

Step 3 — solve the modulus.

k+6=25 \Rightarrow k=19

k+6=-25 \Rightarrow k=-31

Step 4 — write the two tangents.

4x+3y+19=0 \text{and}4x+3y-31=0

These are the two parallel tangents on opposite sides of the circle, which is why one constant is positive and the other negative.

Answer: Option 4

Shortcut. The two parallel tangents sit symmetrically about the centre, so for 4x+3y+k=0,

k=-\left( 4{x}_{0}+3{y}_{0}\right)\pm r\sqrt{{a}^{2}+{b}^{2}}=-6\pm 25

giving 19 and -31 at once — no modulus case work.

Use that symmetry to eliminate. The mean of the two constants must be -6:

One average, one answer. Note also the 3–4–5 structure: \sqrt{{4}^{2}+{3}^{2}}=5=r, which is why the arithmetic is so clean.