Let a variable line passing through the centre of the circle x^2 + y^2 - 16x - 4y = 0$x^2 + y^2 - 16x - 4y = 0$, meet the positive co-ordinate axes at the point A$A$ and B$B$. Then the minimum value of OA + OB$OA + OB$, where O$O$ is the origin, is equal to
A.12$12$
B.18$18$
C.20$20$
D.24$24$
Solution & Explanation
### Related Formula
textIntercept form of line: fracxa + fracyb = 1$\text{Intercept form of line: } \frac{x}{a} + \frac{y}{b} = 1$
### Core Logic
Circle x^2 + y^2 - 16x - 4y = 0$x^2 + y^2 - 16x - 4y = 0$ has its centre at (8, 2)$(8, 2)$.
Let the line passing through (8, 2)$(8, 2)$ have slope m$m$. Its equation is:
y - 2 = m(x - 8)$y - 2 = m(x - 8)$
x-intercept (A$A$): set y=0 implies -2 = m(x-8) implies x = 8 - frac2m$y=0 \implies -2 = m(x-8) \implies x = 8 - \frac{2}{m}$.
y-intercept (B$B$): set x=0 implies y = 2 - 8m$x=0 \implies y = 2 - 8m$.
Sum of intercepts OA + OB = (8 - frac2m) + (2 - 8m) = 10 - frac2m - 8m$OA + OB = (8 - \frac{2}{m}) + (2 - 8m) = 10 - \frac{2}{m} - 8m$.
To minimize, let f(m) = 10 - frac2m - 8m$f(m) = 10 - \frac{2}{m} - 8m$.
f'(m) = frac2m^2 - 8 = 0 implies m^2 = frac14$f'(m) = \frac{2}{m^2} - 8 = 0 \implies m^2 = \frac{1}{4}$
Since the line meets the positive coordinate axes, intercepts must be positive, which requires m < 0$m < 0$. Thus m = -1/2$m = -1/2$.
Substitute m = -1/2$m = -1/2$:
OA + OB = 10 - frac2-1/2 - 8(-1/2) = 10 + 4 + 4 = 18$OA + OB = 10 - \frac{2}{-1/2} - 8(-1/2) = 10 + 4 + 4 = 18$
### Pattern Recognition
AM-GM can also be applied: 8a + 2b = ab implies 1 = frac8a + frac2b$8a + 2b = ab \implies 1 = \frac{8}{a} + \frac{2}{b}$. To minimize a+b$a+b$, use Cauchy-Schwarz or standard differentiation. Differentiation directly yields intercept minima.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Circles
Keywords:#variable line#JEE Main 2024 Evening Q4#Circles JEE Main 2024#Equation of Tangent and Normal JEE Main 2024
More Circles Previous-Year Questions — Page 3
Q4jee_main_2024_31_jan_morningIntersection and Common Chords
If one of the diameters of the circle x^2 + y^2 - 10x + 4y + 13 = 0$x^2 + y^2 - 10x + 4y + 13 = 0$ is a chord of another circle C, whose center is the point of intersection of the lines 2x + 3y = 12$2x + 3y = 12$ and 3x - 2y = 5$3x - 2y = 5$, then the radius of the circle C is
A.sqrt20$\sqrt{20}$
B.4$4$
C.6$6$
D.3sqrt2$3\sqrt{2}$
Solution
### Core Logic
Find the center of circle C by solving 2x + 3y = 12$2x + 3y = 12$ and 3x - 2y = 5$3x - 2y = 5$.
Multiplying and subtracting yields 13x = 39 implies x = 3, y = 2$13x = 39 \implies x = 3, y = 2$.
Center of C is (3, 2)$(3, 2)$.
### Step 1: Properties of Given Circle
Given circle: x^2 + y^2 - 10x + 4y + 13 = 0$x^2 + y^2 - 10x + 4y + 13 = 0$.
Center M(5, -2)$M(5, -2)$.
Radius r = sqrt25 + 4 - 13 = 4$r = \sqrt{25 + 4 - 13} = 4$.
Intersection and Common Chords diagram for Q4 - JEE Main 2024 Morning
### Step 2: Radius Calculation
The diameter of the first circle is a chord of circle C. Therefore, the distance between the two centers forms a right-angled triangle with the radius of C (CP$CP$) and the radius of the first circle (r = 4$r = 4$).
Distance CM = sqrt(5-3)^2 + (-2-2)^2 = sqrt4 + 16 = sqrt20$CM = \sqrt{(5-3)^2 + (-2-2)^2} = \sqrt{4 + 16} = \sqrt{20}$.
Radius of circle C is CP = sqrtCM^2 + r^2 = sqrt20 + 16 = sqrt36 = 6$CP = \sqrt{CM^2 + r^2} = \sqrt{20 + 16} = \sqrt{36} = 6$.
### Pattern Recognition
When a diameter of circle 1 is a chord of circle 2, the triangle formed by the centers and the point of intersection is a right-angled triangle at the center of circle 1.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Circles
Class 11 Maths: Straight Lines
More Circles Questions — jee_main_2024_31_jan_evening
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