Without changing the direction of the coordin | vedclass

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

Question

(C) Let the new coordinates be $(x', y')$ such that $x = x' + h$ and $y = y' + k$. Substituting these into the given equation $x^2 + y^2 - 4x + 6y - 7

Vedclass · Accepted Answer

$(2, -3)$

Vedclass · Answer

(C) Let the new coordinates be $(x', y')$ such that $x = x' + h$ and $y = y' + k$. Substituting these into the given equation $x^2 + y^2 - 4x + 6y - 7 = 0$ gives:$(x' + h)^2 + (y' + k)^2 - 4(x' + h) + 6(y' + k) - 7 = 0$Expanding the terms:$x'^2 + 2hx' + h^2 + y'^2 + 2ky' + k^2 - 4x' - 4h + 6y' + 6k - 7 = 0$Grouping the terms:$x'^2 + y'^2 + x'(2h - 4) + y'(2k + 6) + (h^2 + k^2 - 4h + 6k - 7) = 0$To eliminate the linear terms,the coefficients of $x'$ and $y'$ must be zero:$2h - 4 = 0 \Rightarrow h = 2$$2k + 6 = 0 \Rightarrow k = -3$Thus,the point $(h, k)$ is $(2, -3)$.

Without changing the direction of the coordinate axes,the origin is shifted to $(h, k)$ such that the linear (first-degree) terms in the equation $x^2 + y^2 - 4x + 6y - 7 = 0$ are eliminated. Then the point $(h, k)$ is:

Similar Questions

The locus of the point which is equidistant from the point $(1,1)$ and the line $x+y+1=0$ is

The locus of the centre of a circle which passes through the origin and cuts off a length of $4$ units from the line $x=3$ is

Let $S$ be the set of points on the $X$-axis lying at a distance of $d$ units from $(3, 4)$. Which of the following is true?

If a circle of radius $R$ passes through the origin $O$ and intersects the coordinate axes at $A$ and $B,$ then the locus of the foot of the perpendicular from $O$ on $AB$ is

$A$ point moves in such a way that the sum of the squares of its distances from the points $A(2, 0)$ and $B(-2, 0)$ is always equal to the square of the distance between $A$ and $B$. The locus of the point is

Vedclass Test Series

Exam Paper Generator

Online Exam Module