Mathematics - Circle Question with Solution | TestHub

S is orthocentre of $△PQR$.

Problem: A line $L_1$ intersects x and $y$ axes at P and Q respectively. Another line $L_2$ perpendicular to $L_1$ cuts the $x$ and $y$ axes at R and S respectively. The locus of the point of intersection of the lines PS and QR is a circle passing through

Options:

A. The origin

B. P

C. Q

D. R

Correct Answer: A, B, C

Detailed explanation:

Let the equation of line $L_1$ be $\frac{x}{a}+\frac{y}{b}=1$.

This means line $L_1$ intersects the x-axis at $P(a,0)$ and the y-axis at $Q(0,b)$.

The slope of $L_1$ is $m_1=-\frac{b}{a}$.

Line $L_2$ is perpendicular to $L_1$. Therefore, the slope of $L_2$ is $m_2=-\frac{1}{m_1}=-\frac{1}{-\frac{b}{a}}=\frac{a}{b}$.

Let the equation of line $L_2$ be $\frac{x}{c}+\frac{y}{d}=1$.

This means line $L_2$ intersects the x-axis at $R(c,0)$ and the y-axis at $S(0,d)$.

The slope of $L_2$ is also $m_2=-\frac{d}{c}$.

So, we have $\frac{a}{b}=-\frac{d}{c}$, which implies $ac=-bd$.

Now we need to find the locus of the point of intersection of lines PS and QR.

Let the point of intersection be $K(h,k)$.

Line PS passes through $P(a,0)$ and $S(0,d)$.

The equation of line PS is $\frac{x}{a}+\frac{y}{d}=1$.

Since $K(h,k)$ lies on PS, we have $\frac{h}{a}+\frac{k}{d}=1$ (Equation 1)

Line QR passes through $Q(0,b)$ and $R(c,0)$.

The equation of line QR is $\frac{x}{c}+\frac{y}{b}=1$.

Since $K(h,k)$ lies on QR, we have $\frac{h}{c}+\frac{k}{b}=1$ (Equation 2)

From Equation 1: $dh+ak=ad$

From Equation 2: $bh+ck=bc$

We have the condition $ac=-bd$.

Let's express $c$ in terms of $a,b,d$: $c=-\frac{bd}{a}$.

Substitute $c$ into Equation 2:

$\frac{h}{-\frac{bd}{a}}+\frac{k}{b}=1$

$-\frac{ah}{bd}+\frac{k}{b}=1$

Multiply by $bd$: $-ah+dk=bd$ (Equation 3)

Now we have two equations for $h$ and $k$:

1) $dh+ak=ad$

3) $-ah+dk=bd$

Multiply Equation 1 by $a$ and Equation 3 by $d$:

$adh+a^{2}k=a^{2}d$

$-adh+d^{2}k=b{d}^2$

Add these two equations:

$(a^2+d^2)k=a^{2}d+b{d}^2=d(a^2+bd)$

$k=\frac{d(a^2+bd)}{a^2+d^2}$

Multiply Equation 1 by $d$ and Equation 3 by $a$:

$d^{2}h+adk=a{d}^2$

$-a^{2}h+adk=abd$

Subtract the second from the first:

$(d^2+a^2)h=a{d}^2-abd=a(d^2-bd)$

$h=\frac{a(d^2-bd)}{a^2+d^2}$

This approach seems complicated. Let's try a different geometric interpretation.

Consider the quadrilateral PRQS.

P is $(a,0)$, Q is $(0,b)$, R is $(c,0)$, S is $(0,d)$.

The lines $L_1$ (PQ) and $L_2$ (RS) are perpendicular.

The x-axis is PR. The y-axis is QS.

Let $K(h,k)$ be the intersection of PS and QR.

Consider the triangle formed by the origin O(0,0), P(a,0), and Q(0,b).

The line $L_1$ is PQ.

The line $L_2$ is RS.

The x-axis is the line PR. The y-axis is the line QS.

The slope of PQ is $m_{PQ}=\frac{b-0}{0-a}=-\frac{b}{a}$.

The slope of RS is $m_{RS}=\frac{d-0}{0-c}=-\frac{d}{c}$.

Since $L_1\perp L_2$, $m_{PQ}\cdot m_{RS}=-1$.

$(-\frac{b}{a})\cdot (-\frac{d}{c})=-1$

$\frac{bd}{ac}=-1⟹bd=-ac$.

Now consider the lines PS and QR.

Line PS passes through $P(a,0)$ and $S(0,d)$.

Equation of PS: $\frac{x}{a}+\frac{y}{d}=1$.

Line QR passes through $Q(0,b)$ and $R(c,0)$.

Equation of QR: $\frac{x}{c}+\frac{y}{b}=1$.

Let $(h,k)$ be the point of intersection.

$\frac{h}{a}+\frac{k}{d}=1$ (1)

$\frac{h}{c}+\frac{k}{b}=1$ (2)

From (1), $dh+ak=ad$.

From (2), $bh+ck=bc$.

We know $bd=-ac$.

From (1), $k=\frac{ad-dh}{a}=d-\frac{d}{a}h$.

Substitute into (2):

$\frac{h}{c}+\frac{1}{b}(d-\frac{d}{a}h)=1$

$\frac{h}{c}+\frac{d}{b}-\frac{dh}{ab}=1$

Multiply by $abc$:

$abh+acd-cdh=abc$

$abh-cdh=abc-acd$

$h(ab-cd)=ac(b-d)$

We know $c=-\frac{bd}{a}$.

$ab-cd=ab-(-\frac{bd}{a})d=ab+\frac{b{d}^2}{a}=\frac{a^{2}b+b{d}^2}{a}=\frac{b(a^2+d^2)}{a}$.

$ac(b-d)=a(-\frac{bd}{a})(b-d)=-bd(b-d)$.

So, $h\frac{b(a^2+d^2)}{a}=-bd(b-d)$.

$h\frac{a^2+d^2}{a}=-d(b-d)$.

$h=\frac{-ad(b-d)}{a^2+d^2}=\frac{ad(d-b)}{a^2+d^2}$.

Now for $k$:

From (2), $k=\frac{bc-bh}{c}=b-\frac{b}{c}h$.

Substitute into (1):

$\frac{h}{a}+\frac{1}{d}(b-\frac{b}{c}h)=1$

$\frac{h}{a}+\frac{b}{d}-\frac{bh}{cd}=1$

Multiply by $acd$:

$cdh+abc-abh=acd$

$cdh-abh=acd-abc$

$h(cd-ab)=ac(d-b)$

$h=\frac{ac(d-b)}{cd-ab}$.

This is consistent with the previous calculation for $h$.

Now let's find $k$.

From $dh+ak=ad$, we have $ak=ad-dh$.

$k=d-\frac{d}{a}h=d-\frac{d}{a}\left(\frac{ad(d-b)}{a^2+d^2}\right)$

$k=d-\frac{d^2(d-b)}{a^2+d^2}=\frac{d(a^2+d^2)-d^2(d-b)}{a^2+d^2}$

$k=\frac{a^{2}d+d^3-d^3+b{d}^2}{a^2+d^2}=\frac{a^{2}d+b{d}^2}{a^2+d^2}=\frac{d(a^2+bd)}{a^2+d^2}$.

We have $h=\frac{ad(d-b)}{a^2+d^2}$ and $k=\frac{d(a^2+bd)}{a^2+d^2}$.

We also have the condition $bd=-ac$.

Substitute $bd=-ac$ into the expression for $k$:

$k=\frac{d(a^2-ac)}{a^2+d^2}=\frac{ad(a-c)}{a^2+d^2}$.

Now let's look at $h^2+k^2$. This is not directly leading to a simple circle equation.

Let's consider the angles.

Let the angle of $L_1$ with the x-axis be $\theta$. Then $m_1=\tan \theta =-\frac{b}{a}$.

The angle of $L_2$ with the x-axis is $\theta +\frac{\pi }{2}$. Then $m_2=\tan (\theta +\frac{\pi }{2})=-\cot \theta =\frac{a}{b}$.

We have $m_2=-\frac{d}{c}$, so $\frac{a}{b}=-\frac{d}{c}⟹ac=-bd$.

Consider the quadrilateral PRQS.

P is $(a,0)$, R is $(c,0)$. So PR lies on the x-axis.

Q is $(0,b)$, S is $(0,d)$. So QS lies on the y-axis.

The origin O is $(0,0)$.

The line PS has equation $\frac{x}{a}+\frac{y}{d}=1$.

The line QR has equation $\frac{x}{c}+\frac{y}{b}=1$.

Let the point of intersection be $(x,y)$.

$x/a+y/d=1⟹dx+ay=ad$

$x/c+y/b=1⟹bx+cy=bc$

From $bd=-ac$, we have $c=-bd/a$.

Substitute $c$ into the second equation:

$bx-\frac{bd}{a}y=bc$

Multiply by $a$: $abx-bdy=abc$

Divide by $b$: $ax-dy=ac$

Now we have two equations for $x,y$:

1) $dx+ay=ad$

2) $ax-dy=ac$

Multiply (1) by $d$ and (2) by $a$:

$d^{2}x+ady=a{d}^2$

$a^{2}x-ady=a^{2}c$

Add them:

$(d^2+a^2)x=a{d}^2+a^{2}c$

$x=\frac{a{d}^2+a^{2}c}{a^2+d^2}$

Multiply (1) by $a$ and (2) by $d$:

$adx+a^{2}y=a^{2}d$

$adx-d^{2}y=adc$

Subtract the second from the first:

$(a^2+d^2)y=a^{2}d-adc$

$y=\frac{a^{2}d-adc}{a^2+d^2}$

Now, let's use the condition $ac=-bd$.

$x=\frac{a{d}^2+a(-bd)}{a^2+d^2}=\frac{a{d}^2-abd}{a^2+d^2}=\frac{ad(d-b)}{a^2+d^2}$.

$y=\frac{a^{2}d-a(-bd)}{a^2+d^2}=\frac{a^{2}d+abd}{a^2+d^2}=\frac{ad(a+b)}{a^2+d^2}$.

This is not $y=\frac{d(a^2+bd)}{a^2+d^2}$ as before. Let's recheck.

$y=\frac{a^{2}d-adc}{a^2+d^2}$.

Substitute $c=-bd/a$:

$y=\frac{a^{2}d-ad(-bd/a)}{a^2+d^2}=\frac{a^{2}d+b{d}^2}{a^2+d^2}=\frac{d(a^2+bd)}{a^2+d^2}$. This is correct.

So, the coordinates of the point of intersection $K(x,y)$ are:

$x=\frac{ad(d-b)}{a^2+d^2}$

$y=\frac{d(a^2+bd)}{a^2+d^2}$

We need to find the locus of $(x,y)$.

Consider the product $x^2+y^2$. This will be complicated.

Let's try to find a relation between $x$ and $y$.

From $dx+ay=ad$ and $ax-dy=ac$.

We can eliminate $d$ and $a$.

From $dx+ay=ad⟹d(x-a)=-ay⟹d=\frac{-ay}{x-a}$.

From $ax-dy=ac⟹a(x-c)=dy⟹a=\frac{dy}{x-c}$.

Substitute $d$ into $a$:

$a=\frac{(\frac{-ay}{x-a})y}{x-c}=\frac{-a{y}^2}{(x-a)(x-c)}$.

$1=\frac{-y^2}{(x-a)(x-c)}$

$(x-a)(x-c)=-y^2$

$x^2-(a+c)x+ac=-y^2$

$x^2+y^2-(a+c)x+ac=0$.

This is the equation of a circle.

We know $ac=-bd$. So $x^2+y^2-(a+c)x-bd=0$.

This equation depends on $a,c,b,d$. We need to eliminate them.

Let's use the property that $ac=-bd$.

The equation of the locus is $x^2+y^2-(a+c)x+ac=0$.

This circle passes through the origin $(0,0)$ because $0^2+0^2-(a+c)(0)+ac=ac$.

If $ac=0$, then it passes through the origin.

However, $a,b,c,d$ are non-zero for the lines to intersect at distinct points on the axes.

The equation $x^2+y^2-(a+c)x+ac=0$ is the locus.

Let's check if it passes through P, Q, R, S.

P is $(a,0)$.

$a^2+0^2-(a+c)a+ac=a^2-a^2-ac+ac=0$.

So the circle passes through P.

Q is $(0,b)$.

$0^2+b^2-(a+c)(0)+ac=b^2+ac$.

For Q to be on the circle, $b^2+ac=0$.

We know $ac=-bd$. So $b^2-bd=0⟹b(b-d)=0$.

Since $b\ne 0$, we must have $b=d$.

If $b=d$, then $ac=-b^2$.

In this case, the circle passes through Q.

R is $(c,0)$.

$c^2+0^2-(a+c)c+ac=c^2-ac-c^2+ac=0$.

So the circle passes through R.

S is $(0,d)$.

$0^2+d^2-(a+c)(0)+ac=d^2+ac$.

For S to be on the circle, $d^2+ac=0$.

We know $ac=-bd$. So $d^2-bd=0⟹d(d-b)=0$.

Since $d\ne 0$, we must have $d=b$.

If $d=b$, then the circle passes through S.

The question asks for the locus of the point of intersection of PS and QR.

The equation of the locus is $x^2+y^2-(a+c)x+ac=0$.

This is a circle.

Let's re-examine the condition $bd=-ac$.

This condition is always true for the given problem setup.

The equation of the locus is $x^2+y^2-(a+c)x+ac=0$.

This circle passes through:

1. Origin $(0,0)$ if $ac=0$. But $a,c$ are x-intercepts, so they are generally non-zero.

However, if $ac\ne 0$, the circle does not pass through the origin.

Let's recheck the derivation of the locus.

We have $dx+ay=ad$ and $ax-dy=ac$.

Multiply the first by $x$: $d{x}^2+axy=adx$.

Multiply the second by $y$: $axy-d{y}^2=acy$.

Subtract the second from the first:

$d{x}^2+d{y}^2=adx-acy$

$d(x^2+y^2)=a(dx-cy)$.

This is not leading to the previous equation. Let's re-derive the locus equation carefully.

We have:

1) $dx+ay=ad$

2) $bx+cy=bc$

And $ac=-bd$.

From (1), $d(x-a)=-ay$.

From (2), $c(y-b)=-bx$.

Let's eliminate $d$ and $c$.

$d=\frac{-ay}{x-a}$.

$c=\frac{-bx}{y-b}$.

Substitute $d$ and $c$ into $ac=-bd$:

$a\left(\frac{-bx}{y-b}\right)=-b\left(\frac{-ay}{x-a}\right)$

$\frac{-abx}{y-b}=\frac{aby}{x-a}$

Since $a,b\ne 0$, we can divide by $ab$:

$\frac{-x}{y-b}=\frac{y}{x-a}$

$-x(x-a)=y(y-b)$

$-x^2+ax=y^2-by$

$x^2+y^2-ax-by=0$.

This is the correct equation of the locus.

This is a circle.

Let's check which points it passes through:

1. Origin $(0,0)$: $0^2+0^2-a(0)-b(0)=0$. So it passes through the origin. (Option A is correct)

2. Point P $(a,0)$: $a^2+0^2-a(a)-b(0)=a^2-a^2=0$. So it passes through P. (Option B is correct)

3. Point Q $(0,b)$: $0^2+b^2-a(0)-b(b)=b^2-b^2=0$. So it passes through Q. (Option C is correct)

4. Point R $(c,0)$: $c^2+0^2-a(c)-b(0)=c^2-ac$.

For R to be on the circle, $c^2-ac=0⟹c(c-a)=0$.

Since $c\ne 0$, we must have $c=a$.

If $c=a$, then $ac=-bd⟹a^2=-bd$. This is a specific condition, not generally true.

So, the circle does not necessarily pass through R. (Option D is incorrect)

5. Point S $(0,d)$: $0^2+d^2-a(0)-b(d)=d^2-bd$.

For S to be on the circle, $d^2-bd=0⟹d(d-b)=0$.

Since $d\ne 0$, we must have $d=b$.

If $d=b$, then $ac=-b^2$. This is a specific condition, not generally true.

So, the circle does not necessarily pass through S.

Therefore, the locus of the point of intersection of lines PS and QR is a circle passing through the origin, P, and Q.

The statement in the current solution "S is orthocentre of $△PQR$" is incorrect and irrelevant to the question. The orthocenter of $△PQR$ would be the intersection of altitudes.

Final check of the locus equation $x^2+y^2-ax-by=0$.

This is a circle with center $(\frac{a}{2},\frac{b}{2})$ and radius $\sqrt{(\frac{a}{2}{)}^2+(\frac{b}{2}{)}^2}=\frac{1}{2}\sqrt{a^2+b^2}$.

This circle passes through the origin $(0,0)$, $P(a,0)$, and $Q(0,b)$.

The condition $ac=-bd$ is crucial.

The lines $L_1$ and $L_2$ are perpendicular.

$L_1:\frac{x}{a}+\frac{y}{b}=1$

$L_2:\frac{x}{c}+\frac{y}{d}=1$

Slope of $L_1$ is $-\frac{b}{a}$. Slope of $L_2$ is $-\frac{d}{c}$.

Product of slopes: $(-\frac{b}{a})(-\frac{d}{c})=-1⟹\frac{bd}{ac}=-1⟹bd=-ac$. This is correctly used.

The points are:

$P(a,0)$

$Q(0,b)$

$R(c,0)$

$S(0,d)$

Line PS: $\frac{x}{a}+\frac{y}{d}=1$

Line QR: $\frac{x}{c}+\frac{y}{b}=1$

Let $(x,y)$ be the intersection point.

From $\frac{x}{a}+\frac{y}{d}=1⟹dx+ay=ad$.

From $\frac{x}{c}+\frac{y}{b}=1⟹bx+cy=bc$.

From $dx+ay=ad⟹d(x-a)=-ay$.

From $bx+cy=bc⟹c(y-b)=-bx$.

Substitute $d=\frac{-ay}{x-a}$ and $c=\frac{-bx}{y-b}$ into $bd=-ac$.

$b\left(\frac{-ay}{x-a}\right)=-a\left(\frac{-bx}{y-b}\right)$

$\frac{-aby}{x-a}=\frac{abx}{y-b}$

Since $a,b\ne 0$, we can divide by $ab$:

$\frac{-y}{x-a}=\frac{x}{y-b}$

$-y(y-b)=x(x-a)$

$-y^2+by=x^2-ax$

$x^2+y^2-ax-by=0$.

This is indeed the correct locus equation.

The circle $x^2+y^2-ax-by=0$ passes through:

- Origin $(0,0)$: $0^2+0^2-a(0)-b(0)=0$.

- $P(a,0)$: $a^2+0^2-a(a)-b(0)=a^2-a^2=0$.

- $Q(0,b)$: $0^2+b^2-a(0)-b(b)=b^2-b^2=0$.

Therefore, the locus of the point of intersection of the lines PS and QR is a circle passing through the origin, P, and Q.

The final answer is $\boxed{A,B,C}$.