If a circle passes through the points of inte | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The points of intersection of the line $\lambda x - y + 1 = 0$ with the coordinate axes are $(-\frac{1}{\lambda}, 0)$ and $(0, 1)$.The points of i

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The points of intersection of the line $\lambda x - y + 1 = 0$ with the coordinate axes are $(-\frac{1}{\lambda}, 0)$ and $(0, 1)$.The points of intersection of the line $x - 2y + 3 = 0$ with the coordinate axes are $(-3, 0)$ and $(0, \frac{3}{2})$.Since the circle passes through these four points,it passes through $(0, 1), (-3, 0),$ and $(0, \frac{3}{2})$.The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.Substituting $(0, 1)$,we get $1 + 2f + c = 0$.Substituting $(0, \frac{3}{2})$,we get $\frac{9}{4} + 3f + c = 0$.Subtracting these equations: $(3f - 2f) + (\frac{9}{4} - 1) = 0 \Rightarrow f = -\frac{5}{4}$.Then $c = -1 - 2(-\frac{5}{4}) = -1 + \frac{5}{2} = \frac{3}{2}$.Substituting $(-3, 0)$,we get $9 - 6g + \frac{3}{2} = 0$ $\Rightarrow 6g = \frac{21}{2}$ $\Rightarrow g = \frac{7}{4}$.The circle equation is $x^2 + y^2 + \frac{7}{2}x - \frac{5}{2}y + \frac{3}{2} = 0$.Since it passes through $(-\frac{1}{\lambda}, 0)$,we have $\frac{1}{\lambda^2} - \frac{7}{2\lambda} + \frac{3}{2} = 0$.Multiplying by $2\lambda^2$: $2 - 7\lambda + 3\lambda^2 = 0$.Solving $3\lambda^2 - 7\lambda + 2 = 0$,we get $(3\lambda - 1)(\lambda - 2) = 0$.Thus,$\lambda = 2$ or $\lambda = \frac{1}{3}$. Given the options,the correct value is $2$.

If a circle passes through the points of intersection of the coordinate axes with the lines $\lambda x - y + 1 = 0$ and $x - 2y + 3 = 0$,then the value of $\lambda$ is

Similar Questions

Consider the curves $C_1: y^2=4x$ and $C_2: x^2+y^2-6x+1=0$. Assertion $(A)$: The common tangents to the curves $C_1$ and $C_2$ are orthogonal. Reason $(R)$: $x-y+1=0$ and $x+y+1=0$ are the common tangents to the curves $C_1$ and $C_2$.

For different real non-zero numbers $x_1, x_2, x_3$ and $x_4$,suppose the points $(x_1, \frac{1}{x_1}), (x_2, \frac{1}{x_2}), (x_3, \frac{1}{x_3})$ and $(x_4, \frac{1}{x_4})$ lie on the boundary of a circle of radius $4$. Then,the value of $x_1 x_2 x_3 x_4$ is

The circle $x^2 + y^2 - 8x = 0$ and the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ intersect at points $A$ and $B$. Find the equation of the circle with $AB$ as its diameter.

$A(x_1, y_1)$ is the internal centre of similitude and $B(x_2, y_2)$ is the external centre of similitude of two circles $C_1$ and $C_2$ whose centres are $P(\alpha, \beta)$ and $Q(\gamma, \delta)$ respectively. If $PA=3, AB=5, QB=2$,then the ratio of the radii of the two circles is:

If the ratio of the lengths of tangents drawn from the point $(f, g)$ to the circles $x^2 + y^2 = 6$ and $x^2 + y^2 + 3x + 3y = 0$ is $2 : 1$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module