JEE Main 2024 (Online) 29th January Morning ShiftQuestion 1
Equations of two diameters of a circle are $2x-3y=5$ and $3x-4y=7$. The line joining the points $\left(-\frac{22}{7},-4\right)$ and $\left(-\frac{1}{7},3\right)$ intersects the circle at only one point $P(\alpha,\beta)$. Then $17\beta-\alpha$ is equal to $\_\_\_\_\_\_$ .
Circle
Answer$2$
Solution
Concept. Diameters meet at the centre. A line meeting the circle at one point is a tangent, and the point of contact is the foot of the perpendicular from the centre. Formulas used. (1) Foot of perpendicular from $(x_1,y_1)$ on $ax+by+c=0$: $(x_1,y_1)-\dfrac{ax_1+by_1+c}{a^2+b^2}(a,b)$Step 1 - Centre. $2x-3y=5$, $3x-4y=7$: $6x-9y=15$, $6x-8y=14$; subtract: $y=-1$, then $x=1$. Centre $(1,-1)$. Step 2 - Line. Slope $=\frac{3+4}{-\frac17+\frac{22}7}=\frac{7}{3}$. Through $\big(-\frac17,3\big)$: $3y-9=7x+1$, i.e. $7x-3y+10=0$. Step 3 - Use (1). $7(1)-3(-1)+10=20$, $a^2+b^2=58$: $$P=(1,-1)-\frac{20}{58}(7,-3)=\Big(-\frac{41}{29},\ \frac{1}{29}\Big).$$ Step 4 - Value. $17\beta-\alpha=\frac{17}{29}+\frac{41}{29}=2$. Answer: 2